EDIT: Part 4 added. EDIT2: Second proof of Part 4 added.
1. The answer is no (as long as we are working over a field - of any characteristic, algebraically closed or not). If $k$ is a field and $G$ is a finite group, then the dimension of any irreducible representation $V$ of $G$ over $k$ is $leq left|Gright|$. This is actually obvious: Take any nonzero vector $vin V$; then, $kleft[Gright]v$ is a nontrivial subrepresentation of $V$ of dimension $leqdimleft(kleft[Gright]right)=left|Gright|$. Since our representation $V$ was irreducible, this subrepresentation must be $V$, and hence $dim Vleqleft|Gright|$.
2. Okay, we can do a little bit better: Any irreducible representation $V$ of $G$ has dimension $leqleft|Gright|-1$, unless $G$ is the trivial group. Same proof applies, with one additional step:
If $dim V=left|Gright|$, then the map $kleft[Gright]to V, gmapsto gv$ must be bijective (in fact, it is surjective,
since $kleft[Gright]v=V$, and it therefore must be bijective since $dimleft(kleft[Gright]right)=left|Gright|=dim V$), so it is an isomorphism of representations (since it is $G$-equivariant), and thus $Vcong kleft[Gright]$. But $kleft[Gright]$ is not an irreducible representation, unless $G$ is the trivial group (in fact, it always contains the $1$-dimensional trivial representation).
3. Note that if the base field $k$ is algebraically closed and of characteristic $0$, then we can do much better: In this case, an irreducible representation of $G$ always has dimension $<sqrt{left|Gright|}$ (in fact, in this case, the sum of the squares of the dimensions of all irreducible representations is $left|Gright|$, and one of these representations is the trivial $1$-dimensional one). However, if the base field is not necessarily algebraically closed and of arbitrary characteristic, then the bound $dim Vleq left|Gright|-1$ can be sharp (take cyclic groups).
4. There is a way to improve 2. so that it comes a bit closer to 3.:
Theorem 1. If $V_1$, $V_2$, ..., $V_m$ are $m$ pairwise nonisomorphic irreducible representations of a finite-dimensional algebra $A$ over a field $k$ (not necessarily algebraically closed, not necessarily of characteristic $0$), then $dim V_1+dim V_2+...+dim V_mleqdim A$.
(Of course, if $A$ is the group algebra of some finite group $G$, then $dim A=left|Gright|$, and we get 2. as a consequence.)
First proof of Theorem 1. At first, for every $iinleftlbrace 1,2,...,mrightrbrace$, the (left) representation $V_i^{ast}$ of the algebra $A^{mathrm{op}}$ (this representation is defined by $acdot f=left(vmapsto fleft(avright)right)$ for any $fin V_i^{ast}$ and $ain A$) is irreducible (since $V_i$ is irreducible) and therefore isomorphic to a quotient of the regular (left) representation $A^{mathrm{op}}$ (since we can choose some nonzero $uin V_i^{ast}$, and then the map $A^{mathrm{op}}to V_i^{ast}$ given by $amapsto au$ must be surjective, because its image is a nonzero subrepresentation of $V_i^{ast}$ and therefore equal to $V_i^{ast}$ due to the irreducibility of $V_i^{ast}$). Hence, by duality, $V_i$ is isomorphic to a subrepresentation of the (left) representation $A^{mathrm{op}ast}=A^{ast}$ of $A$. Hence, from now on, let's assume that $V_i$ actually is a subrepresentation of $A^{ast}$ for every $iinleftlbrace 1,2,...,mrightrbrace$.
Now, let us prove that the vector subspaces $V_1$, $V_2$, ..., $V_m$ of $A^{ast}$ are linearly disjoint, i. e., that the sum $V_1+V_2+...+V_m$ is actually a direct sum. We will prove this by induction over $m$, so let's assume that the sum $V_1+V_2+...+V_{m-1}$ is already a direct sum. It remains to prove that $V_mcap left(V_1+V_2+...+V_{m-1}right)=0$. In fact, assume the contrary. Then, $V_mcap left(V_1+V_2+...+V_{m-1}right)=V_m$ (since $V_mcap left(V_1+V_2+...+V_{m-1}right)$ is a nonzero subrepresentation of $V_m$, and $V_m$ is irreducible). Thus, $V_msubseteq V_1+V_2+...+V_{m-1}$. Consequently, $V_m$ is isomorphic to a subrepresentation of the direct sum $V_1oplus V_2oplus ...oplus V_{m-1}$ (because the sum $V_1+V_2+...+V_{m-1}$ is a direct sum, according to our induction assumption).
Now, according to Theorem 2.2 and Remark 2.3 of Etingof's "Introduction to representation theory", any subrepresentation of the direct sum $V_1oplus V_2oplus ...oplus V_{m-1}$ must be a direct sum of the form $r_1V_1oplus r_2V_2oplus ...oplus r_{m-1}V_{m-1}$ for some nonnegative integers $r_1$, $r_2$, ..., $r_{m-1}$. Hence, every irreducible subrepresentation of the direct sum $V_1oplus V_2oplus ...oplus V_{m-1}$ must be one of the representations $V_1$, $V_2$, ..., $V_{m-1}$. Since we know that $V_m$ is isomorphic to a subrepresentation of the direct sum $V_1oplus V_2oplus ...oplus V_{m-1}$, we conclude that $V_m$ is isomorphic to one of the representations $V_1$, $V_2$, ..., $V_{m-1}$. This contradicts the non-isomorphy of the representations $V_1$, $V_2$, ..., $V_m$. Thus, we have proven that the sum $V_1+V_2+...+V_m$ is actually a direct sum. Consequently, $dim V_1+dim V_2+...+dim V_m=dimleft(V_1+V_2+...+V_mright)leq dim A^{ast}=dim A$, and Theorem 1 is proven.
Second proof of Theorem 1. I just learnt the following simpler proof of Theorem 1 from §1 Lemma 1 in Crawley-Boevey's "Lectures on representation theory and invariant theory":
Let $0=A_0subseteq A_1subseteq A_2subseteq ...subseteq A_k=A$ be a composition series of the regular representation $A$ of $A$. Then, by the definition of a composition series, for every $iin leftlbrace 1,2,...,krightrbrace$, the representation $A_i/A_{i-1}$ of $A$ is irreducible.
Let $T$ be an irreducible representation of $A$. We are going to prove that there exists some $Iin leftlbrace 1,2,...,krightrbrace$ such that $Tcong A_I/A_{I-1}$ (as representations of $A$).
In fact, let $I$ be the smallest element $iin leftlbrace 1,2,...,krightrbrace$ satisfying $A_iTneq 0$ (such elements $i$ exist, because $A_kT=AT=Tneq 0$). Then, $A_ITneq 0$, but $A_{I-1}T=0$. Now, choose some vector $tin T$ such that $A_Itneq 0$ (such a vector $t$ exists, because $A_ITneq 0$), and consider the map $f:A_Ito T$ defined by $fleft(aright)=at$ for every $ain A_I$. Then, this map $f$ is a homomorphism of representations of $A$. Since it maps the subrepresentation $A_{I-1}$ to $0$ (because $fleft(A_{I-1}right)=A_{I-1}tsubseteq A_{I-1}T=0$), it gives rise to a map $g:A_I/A_{I-1}to T$, which, of course, must also be a homomorphism of representations of $A$. Since $A_I/A_{I-1}$ and $T$ are irreducible representations of $A$, it follows from Schur's lemma that any homomorphism of representations from $A_I/A_{I-1}$ to $T$ is either an isomorphism or identically zero. Hence, $g$ is either an isomorphism or identically zero. But $g$ is not identically zero (since $gleft(A_I/A_{I-1}right)=fleft(A_Iright)=A_Itneq 0$), so that $g$ must be an isomorphism, i. e., we have $Tcong A_I/A_{I-1}$.
So we have just proven that
(1) For every irreducible representation $T$ of $A$, there exists some $Iin leftlbrace 1,2,...,krightrbrace$ such that $Tcong A_I/A_{I-1}$ (as representations of $A$).
Denote this $I$ by $I_T$ in order to make it clear that it depends on $T$. So we have $Tcong A_{I_T}/A_{I_T-1}$ for each irreducible representation $T$ of $A$. Applying this to $T=V_i$ for every $iinleftlbrace 1,2,...,mrightrbrace$, we see that $V_icong A_{I_{V_i}}/A_{I_{V_i}-1}$ for every $iinleftlbrace 1,2,...,mrightrbrace$. Hence, the elements $I_{V_1}$, $I_{V_2}$, ..., $I_{V_m}$ of the set $leftlbrace 1,2,...,krightrbrace$ are pairwise distinct (because $I_{V_i}=I_{V_j}$ would yield $V_icong A_{I_{V_i}}/A_{I_{V_i}-1}=A_{I_{V_j}}/A_{I_{V_j}-1}cong V_j$, but the representations $V_1$, $V_2$, ..., $V_m$ are pairwise nonisomorphic), and thus
$sumlimits_{i=1}^{m}dimleft(A_{I_{V_i}}/A_{I_{V_i}-1}right)=sumlimits_{substack{jinleftlbrace 1,2,...,krightrbrace ; \ text{there exists }\ iinleftlbrace 1,2,...,mrightrbrace \ text{ such that }j=I_{V_i}}}dimleft(A_j/A_{j-1}right)$
$leq sumlimits_{jinleftlbrace 1,2,...,krightrbrace}dimleft(A_j/A_{j-1}right)$ (since $dimleft(A_j/A_{j-1}right)geq 0$ for every $j$, so that adding more summands cannot decrease the sum)
$=sumlimits_{j=1}^{k}dimleft(A_j/A_{j-1}right)=sumlimits_{j=1}^{k}left(dim A_j-dim A_{j-1}right)$.
Since $dimleft(A_{I_{V_i}}/A_{I_{V_i}-1}right)=dim V_i$ for each $i$ (due to $A_{I_{V_i}}/A_{I_{V_i}-1}cong V_i$) and $sumlimits_{j=1}^{k}left(dim A_j-dim A_{j-1}right)=dim A$ (in fact, the sum $sumlimits_{j=1}^{k}left(dim A_j-dim A_{j-1}right)$ is a telescopic sum and simplifies to $dim A_k-dim A_0=dim A-dim 0=dim A-0=dim A$), this inequality becomes $sumlimits_{i=1}^{m}dim V_ileqdim A$. This proves Theorem 1.
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