What is the adjoint representation of $\operatorname{GL}_n(\mathbb{R})$?

I would have guessed this was a duplicate, but the closest match I could find was: Definition of differential of Adjoint representation of Lie Group.

In any case you're almost there:

Hint As is common in this context, denote $\mathfrak{gl}(n, \Bbb R) := T_I \operatorname{GL}(n, \Bbb R) \cong \operatorname{Mat}_n(\Bbb R)$; since $\operatorname{GL}(n, \Bbb R)$ is a matrix group, I've used the symbol $I$ for its identity element instead of the generic symbol $e$.

As you write, under the identifications you made, the adjoint representation of $\operatorname{GL}(n, \Bbb R)$ is $$\operatorname{Ad} : \operatorname{GL}(n, \Bbb R) \to \operatorname{Aut}(\mathfrak{gl}(n, \Bbb R)), \qquad \operatorname{Ad}(A)(X) = AXA^{-1} .$$ Then, by definition, to compute the adjoint representation $$\operatorname{ad}: \mathfrak{gl}(n, \Bbb R) \to \operatorname{End}(\mathfrak{gl}(n, \Bbb R)) $$ of $\mathfrak{gl}(n, \Bbb R)$ we just differentiate $\operatorname{Ad}(A)(X)$ with respect to $A$.

So, fix a tangent vector $B \in \mathfrak{gl}(n, \Bbb R) \cong \operatorname{Mat}_n(\Bbb R)$, substitute the expression $I + t B$ for $A$ (which defines a path $t \mapsto I + t B$ in $\operatorname{GL}(n, \Bbb R)$ with tangent vector $B$ at $t = 0$) in $\operatorname{Ad}(A)(X) = AXA^{-1}$, and differentiate with respect to $t$ at $t = 0$.

\begin{align}\operatorname{ad}(I + t B)(X) &= \left.\frac{d}{dt}\right\vert_{t = 0} \operatorname{Ad}(I + t B)(X) \\ &= \left.\frac{d}{dt}\right\vert_{t = 0} [(I + t B) X (I + t B)^{-1}] \\ &= (B) X (I) + (I) X \left.\frac{d}{dt}\right\vert_{t = 0}[(I + t B)^{-1}] .\end{align} Differentiating $(I + t B) (I + t B)^{-1} = I$ and evaluating at $t = 0$ gives that $\left.\frac{d}{dt}\right\vert_{t = 0} [(I + t B)^{-1}] = -B$, and so substituting gives $$\color{#df0000}{\boxed{\operatorname{ad}(B)(X) = B X - X B}} .$$ Under the identifications you've made, this expression is the commutator $[B, X]$ in the ring $\operatorname{Mat}_n(\Bbb R)$.