To make the question a little less open-ended while (I hope) retaining its spirit, let me interpret the question as asking for the rate of growth of the function $mu(k)$ defined as the maximum of $M(w)$ over all nontrivial words $w$ of length up to $k$ in any number of symbols, where length is the number of symbols and their inverses multiplied together in $w$. (E.g., $x^5 y^{-3}$ has length $8$.) George Lowther observed that $M(x^{n!})>n$, so $mu(n!)>n$. One can replace $n!$ by $operatorname{LCM}(1,2,ldots,n)$, which is $e^{(1+o(1))n}$, so this gives $mu(k) > (1-o(1)) log k$ as $k to infty$.
I will improve this by showing that $mu(k)$ is at least of order $k^{1/4}$.
Let $C_2(x,y):=[x,y]=xyx^{-1}y^{-1}$. If $C_N$ has been defined, define $$C_{2N}(x_1,ldots,x_N,y_1,ldots,y_N):=[C_N(x_1,ldots,x_N),C_N(y_1,ldots,y_N)].$$ By induction, if $N$ is a power of $2$, then $C_N$ is a word of length $N^2$ that evaluates to $1$ whenever any of its arguments is $1$.
Given $m ge 1$, let $N$ be the smallest power of $2$ such that $N ge 2 binom{m}{2}$. Construct $w$ by applying $C_N$ to a sequence of arguments including $x_i x_j^{-1}$ for $1 le i < j le m$ and extra distinct interdeterminates inserted so that no two of the $x_i x_j^{-1}$ are adjacent arguments of $C_N$. The extra indeterminates ensure that $w$ is not the trivial word. If $w$ is evaluated on elements of a group of size less than $m$, then by the pigeonhole principle two of the $x_i$ have the same value, so some $x_i x_j^{-1}$ is $1$, so $w$ evaluates to $1$. Thus $M(w)>m$. The length of $w$ is at most $2N^2$, which is of order $m^4$. Thus $mu(k)$ is at least of order $k^{1/4}$.
I have a feeling that this is not best possible$ldots$
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