I'll replace $p$ with $q$ and $i$ with $m$, and make the dependence on $q$ explicit, writing $f_q(n)$. If $n = \prod p_j^{k_j}$ is the prime factorization of $n$, then $n$ divides some integer iff $p_j^{k_j}$ divides it for each $j$. Together with the fact that each term of the sequence $|GL_m(\mathbb{F}_q)|$ divides the next, this gives
$$f_q(n) = \max_j f_q(p_j^{k_j})$$
which reduces the problem to the prime power case. This case is equivalent to studying the $p$-adic valuations $\nu_p(|GL_m(\mathbb{F}_q)|)$, since
$$f_q(p^k) = \min \{ m : \nu_p(|GL_m(\mathbb{F}_q)|) \ge k \}.$$
These $p$-adic valuations are analogous to the $p$-adic valuations $\nu_p(n!)$ of the factorials given by Legendre's formula, and we can write down a version of Legendre's formula here using lifting the exponent (LTE). The case $p \mid q$ is special; we get
$$\nu_p(|GL_m(\mathbb{F}_q)|) = {m \choose 2} \nu_p(q).$$
Otherwise, let $d = \text{ord}_p(q)$ be the smallest $d$ such that $q^d \equiv 1 \bmod p$. Then $p \mid q^k - 1$ iff $d \mid k$, and if $p$ is odd then LTE gives
$$\nu_p(q^{dk} - 1) = \nu_p(q^d - 1) + \nu_p(k)$$
which gives
$$\begin{align*} \nu_p(|GL_m(\mathbb{F}_q)|) &= \sum_{k=1}^{\lfloor \frac{m}{d} \rfloor} \nu_p(q^{dk} - 1) \\
&= \sum_{k=1}^{\lfloor \frac{m}{d} \rfloor} \left( \nu_p(q^d - 1) + \nu_p(k) \right) \\
&= \boxed{ \left\lfloor \frac{m}{d} \right\rfloor \nu_p(q^d - 1) + \nu_p \left( \left\lfloor \frac{m}{d} \right\rfloor ! \right). } \end{align*}$$
There is a similar but slightly more annoying result for the case $p = 2$; in this case LTE splits into cases.
Conceptually we can think about this $p$-adic valuation as telling us the order of the $p$-Sylow subgroup of $GL_m(\mathbb{F}_q)$. The first line of this calculation implies that the $p$-Sylow is the $p$-Sylow of $GL_{\lfloor \frac{m}{d} \rfloor}(\mathbb{F}_{q^d})$, where $\mathbb{F}_{q^d}$ is the smallest extension of $\mathbb{F}_q$ containing an element of order $p$. The third line implies that the $p$-Sylow is the $p$-Sylow of the wreath product $\mathbb{F}_{q^d}^{\times} \wr S_{\lfloor \frac{m}{d} \rfloor}$, which is the wreath product of the $p$-Sylows of its constituents.
In terms of understanding the growth rate of $f$, it's clear that the behavior of this $p$-adic valuation depends very strongly on the order $d = \text{ord}_p(q)$ of $q \bmod p$. At one extreme, if $p \mid q - 1$ then $d = 1$ and $\nu_p(|GL_m(\mathbb{F}_q)|)$ grows like $m \, \nu_p(q - 1) + \frac{m}{p-1}$. At the other extreme, if $q$ is a primitive root $\bmod p$ (as Dermot Craddock points out in the comments) then $d = p - 1$ and $\nu_p(|GL_m(\mathbb{F}_p)|)$ is zero until $m = p - 1$. This is the worst case for upper bounding $f(n)$ with no hypotheses on $n$, also as Dermot points out. As an important special case of our discussion above, if $n = p$ is prime and doesn't divide $q$ then we just have
$$f_q(p) = \text{ord}_p(q) = d$$
and we don't have a better upper bound on this than $p - 1$ in general.
Here is a small example to clarify how to actually use these results to calculate $f(n)$. Since we didn't handle the case $p = 2$ I will take $n$ to be odd, so let's calculate $f_2(225)$, where $225 = 3^2 5^2$ (so $q = 2$). Starting with $p = 3$, we want to calculate
$$d_3 = \text{ord}_3(2) = 2$$
(corresponding to $2^2 \equiv 1 \bmod 3$), which gives
$$\nu_3(|GL_m(\mathbb{F}_2)| = \left\lfloor \frac{m}{2} \right\rfloor + \nu_3 \left( \left\lfloor \frac{m}{2} \right\rfloor ! \right)$$
and we want to know when this first takes on a value at least $2$ to calculate $f(3^2)$. This happens when $m = 4$ (the second term doesn't end up taking on a positive value), so $\boxed{ f_2(9) = 4 }$. Next we take $p = 5$, and we want to calculate
$$d_5 = \text{ord}_5(2) = 4$$
(corresponding to $2^4 \equiv 1 \bmod 5$), which gives
$$\nu_5(|GL_m(\mathbb{F}_2)|) = \left\lfloor \frac{m}{4} \right\rfloor + \nu_5 \left( \left\lfloor \frac{m}{4} \right\rfloor ! \right)$$
and we want to know when this first takes on a value at least $2$. This happens when $m = 8$ (again the second term doesn't get to take on a positive value), so $\boxed{ f_2(25) = 8 }$. This gives
$$\boxed{ f_2(225) = \text{max}(f_2(9), f_2(25)) = \text{max}(4, 8) = 8 }.$$
In general it's not hard to see that we have $f_q(p^k) \le k \, \text{ord}_p(q) \le k(p-1)$ (if $\gcd(p, q) = 1$), and in the above cases this was sharp. Writing $n = \prod p_j^{k_j}$ this gives a general upper bound
$$\begin{align*} f_q(n) &\le \max_j \left( k_j \, \text{ord}_{p_j}(q) \right) \\
&\le \boxed{ \max_j \left( k_j(p_j - 1) \right) } \end{align*}$$
which in the above case was also sharp, again assuming $\gcd(n, q) = 1$. The same bound almost always holds when $\gcd(n, q) \neq 1$ with minor exceptions, e.g. $n = q = 2$.
$f_2(n)$ does not appear to be in the OEIS, but the analogous sequence involving dividing factorials is A002034, the Kempner function. Here's a table of the first few values of $f_2(n)$ (there are no OEIS results for the terms up to $f_2(10)$):
| $n$ |
$f_2(n)$ |
| $2$ |
$2$ |
| $3$ |
$2$ |
| $4$ |
$3$ |
| $5$ |
$4$ |
| $6$ |
$2$ |
| $7$ |
$3$ |
| $8$ |
$3$ |
| $9$ |
$4$ |
| $10$ |
$4$ |
| $11$ |
$10$ |
