Longest subgroup chain in $\mathrm{SL}_n(\mathbb{F}_p)$? What is the length $\ell$ of longest chain of subgroups
$$\{e\} \lneq H_1 \lneq \dotsc \lneq H_\ell = G$$
in  $\mathrm{SL}_n(\mathbb{F}_p)$?
 A: We write $\ell(G)$ for the length of the longest subgroup chain of a group $G$. If $G$ is of Lie type, write $B$ for a Borel subgroup of $G$, and note that $\ell(B)$ is equal to the number of prime divisors of $|B|$ (since $B$ is solvable).
Let me explain the situation in the literature as I understand it. Unfortunately it will only yield a partial answer to the question. We start with this paper....
Solomon, Ron; Turull, Alexandre, Chains of subgroups in groups of Lie type. III, J. Lond. Math. Soc., II. Ser. 44, No. 3, 437-444 (1991). ZBL0776.20007.
... whose main theorem is as follows:
Theorem: For all primes p, there exist an integer $F$ such that if $G=G_r(p^m)$ is a quasisimple of Lie type and $m\geq F$, then $\ell(G)=\ell(B)+r$.
Unfortunately, the requirement that $m\geq F$ means that the result does not apply here... However, one of the main theorems of this paper
Solomon, Ron; Turull, Alexandre, Chains of subgroups in groups of Lie type. I, J. Algebra 132, No. 1, 174-184 (1990). ZBL0714.20011.
implies the following
Theorem: Suppose that $G_r(2^m)$ is a quasisimple group of Lie type. Then $\ell(G)=\ell(B)+r$.
So that, at least, answers your question for $p=2$. For $p\neq 2$, the middle paper in the above series (which includes Seitz as an author) is the following:
Seitz, Gary M.; Solomon, Ron; Turull, Alexandre, Chains of subgroups in groups of Lie type. II, J. Lond. Math. Soc., II. Ser. 42, No. 1, 93-100 (1990). ZBL0728.20018.
Theorem B of that paper implies the following:
Theorem: Suppose that $K$ is a positive real number, $m$ a positive real number and $G_r$ any group scheme of Lie type. Then there exists $p$ a prime such that
$\ell(G)>\ell(B)+r+K$.
Thus the bound for $p=2$ fails "as badly as possible" for odd $p$. I've scanned this paper, and the way they show this theorem for $SL_n(p^m)$ is by looking at the normalizer, $N$, of a torus $T$. 
Edit: I wrote earlier that  one should consider the normalizer of a maximal split torus to see why the latter theorem works. In actual fact one should use the normalizer of non-split torus to get a longer chain. So, for instance, if $n$ is odd and if $p^m+1$ has many more prime divisors than $p^m-1$, then one might look $N$, the normalizer of the torus $(p^{2m}-1)^d$ where $2d+1=n$. One can see that
$$\ell(N) = d\pi(p^{2m}-1) + \ell(S_d),$$
which, depending on $p$ and $n$, will get you a longer chain. To calculate precisely how long a chain you obtain you can use the fact that $\ell(S_n)=\lfloor{\frac{3n-1}{2}}\rfloor - b_n$, where $b_n$ is the number ones in the base $2$ expansion of $n$, by a theorem here:
Cameron, Peter J.; Solomon, Ron; Turull, Alexandre, Chains of subgroups in symmetric groups, J. Algebra 127, No. 2, 340-352 (1989). ZBL0683.20004.
Speculation: Let me conjecture what might be the actual answer to your question. It's easier to work inside $GL_n(p)$ rather than $SL_n(p)$. It seems to me that one can do even better than torus normalizers, by looking at Levi subgroups. Consider a Levi subgroup,
$$L=GL_{d_1}(p^{i_1}) \times \cdots \times GL_{d_k}(p^{i_k}),$$
where $\sum d_j i_j=n$. By taking the Borel subgroup of this, one can obtain a subgroup chain in $L$ of length
$$ f(L)= n-k + \sum\limits_{j=1}^k \left( d_i\pi(p^{i_j}-1) + \frac12d_j(d_j+1)\right).$$
This would then yield a subgroup chain in $G$ of length something like $g(L)=f(L)+k-1+\frac12\left(n^2-\sum_{i=1}^k d_i^2\right)$ (I think). Now $\ell(G)$ would be the maximum of all such values (so maximise over $L$, where we think of $L$ as a partition of the integer $n$).
