Yes, the idea that works for ${\rm GL}(2,q)$ with Abelian subgroups works with higher derived lengths for other ${\rm GL}(N, q).$ If $p$ is a prime which does not divide $q$ then the Sylow $p$-subgroups of ${\rm GL}(n,q)$ are monomial (up to equivalence) over some extension field. An easy indction argument (no pun intended) show that a monomial $p$ goup with a faithful representation of degree $p^{k}$ or less has derived length at most $k+1.$ Hence the Sylow $p$-subgroups of ${\rm GL}(N,q)$ have derived length at most $1 +\log_{p}(N)$ for all such $p.$ If $q$ is a power of $p,$ the bound on the derived length of a Sylow $p$-subgroup of ${\rm GL}(N,q)$ is similar. The nilpotence class of the upper unitriangular group is at most $N-1$, so the derived length is at most $\log_{2}(N),$ because for any nilpotent group $U,$ we have $U^{(k)} \leq L_{2^{k}}(U),$ where $U^{(k)}$ is the $k$-the term of the derived series and $L_{m}(U)$ is the $m$-th term of the lower central series. Hence  in particular, no $p$-group of derived length $2 + \log_2(N)$ or more is a subgroup of any ${\rm GL}(N,q)$.