Let b(G) be the maximum degree of an irreducible character of the finite group G.
This answer is mostly to address the solvable, but not nilpotent case. I already up-voted the other two excellent answers: Jim Humphreys mentioned how the groups of Lie type behave (showing that KG≈q2N can be enormous compared to b(G)≈qN), and Bugs Bunny gave the very clear example of the family of extraspecial p-groups showing the opposite extreme (with KG = p(2n+1)/(p(2n)+2) < p, and b(G) = pn).
One idea is to bound b(G) using important characteristic subgroups:
For any finite group G, b(G) ≤ √[G:Z(G)] and b(G) divides [G:Z(G)].
If G is solvable, then Gluck's conjecture is that √[G:Fit(G)] ≤ b(G), and this has been verified for solvable G such that G/Φ(G) has an Abelian Sylow 2-subgroup or G such that G″ = 1. (If G is non-abelian simple, then Fit(G)=1, and so the bound cannot hold).
Another idea is to bound b(G) using abelian subgroups:
If A is an abelian subgroup of G, then b(G) ≤ [G:A]. If A is abelian and subnormal, then in fact b(G) divides [G:A].
If G is solvable, then min(√lg([G:A])) ≤ lg(b(G)). If G is solvable of odd order, then min([G:A]1/6) ≤ b(G). If γ∞(G) is abelian, then min([G:A]1/4) ≤ b(G).
Versions of most of these results are in Isaacs's book (page 28, 30, 84, 190, 212, 216), but I tried to include the 21st century versions where I found them.
I would also caution against thinking "having an abelian subgroup of small index" as "close to abelian". Perhaps there is some truth to it, but the p-groups that are furthest from being abelian (maximal class, that is, having the largest nilpotency class possible amongst groups of the same order) also tend to have the largest abelian subgroups, often index p. Poor extra-special groups are nilpotency class 2 and nearly abelian (they are understood as symplectic vector spaces), yet their abelian subgroups are tiny, |A| ≤ p√[G:Z]. In other words extremely not-abelian groups may have large abelian subgroups, and extremely abelian groups may not.
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