Just to flesh out Geoff's comment....
You can (more or less) findcalculate the exact length of a maximal subgroup chain in $GL_n(q)$ by consideringas a function of the length of a subgroup chain of a Borel subgroup. TheSince the length of thisthe latter chain is easy to calculate - it(it is a function of $\log_pq$ and the prime factorization of $q-1$), one can obtain an exact solution to the question you ask.
To make this comment precise, refer to this paper:
Solomon, Ron; Turull, Alexandre, Chains of subgroups in groups of Lie type. III. J. London Math. Soc. (2) 44 (1991), no. 3, 437–444.
The main result states:
Theorem A∗: Let $p$ be a prime. There exists a positive integer $F(p)$ such that whenever $G=G_r(k)$ is a finite quasisimple group of Lie type with $|k|=p^m$ and $m\geq F(p)$, then $l(G)=l(B)+r$, where $B$ is a Borel subgroup of $G$; moreover, every chain in $G$ of maximal length includes a maximal parabolic subgroup.
(Here we write $l(G)$ for the length of a subgroup chain in a group $G$.) Of course $GL_r(k)$ is not quasisimple, but $SL_r(k)$$SL_{r+1}(k)$ is (unless $r$ and $k$ are very small), soand it is easy to conclude that the result givessame statement holds for $GL_{r+1}(k)$.
Note that the info you need$r$ in the formula corresponds to the number of fundamental roots in the root system corresponding to $G$, and hence corresponds to the length of any maximal chain of parabolic subgroups (the minimal element of which must be a Borel). Thus it is easy to exhibit a subgroup chain of maximal length.
The other two papers in this series are also worth a look - the second has an additional author, Gary Seitz.