Let's try the Baker–Campbell–Hausdorff formula in the form
$$
e^X Y e^{-X} = Y + [X, Y] + \frac{1}{2!}[X,[X,Y]] + \frac{1}{3!}[X,[X,[X,Y]]]\; + \;...
$$
Take $X = ix_jA_{jk}p_k$ and $Y = x_i$. It is not the simplest case since the first commutator,
$$
[X, Y] = [ix_jA_{jk}p_k, x_i] = ix_jA_{jk}(-i\delta_{ki}) = x_jA_{ji}
$$
does not commute with $X = ix_jA_{jk}p_k$. But, let's look at the higher order commutators:
$$
[X, [X, Y]] = [ix_jA_{jk}p_k, x_{j'}A_{j'i}] = ix_jA_{jk}[p_k, x_{j'}]A_{j'i} = ix_jA_{jk} (-i \delta_{kj'})A_{j'i} = x_jA_{jk}A_{ki}
$$
Similarly,
$$
[X, [X, [X, Y]]] =x_jA_{jk}A_{kl}A_{li}\;\; \text{etc.}
$$
Let's take everything in the BCH formula:
$$
e^{ix_jA_{jk}p_k} x_i e^{-ix_jA_{jk}p_k} = x_j\delta_{ji} + x_jA_{ji} + \frac{1}{2!} x_jA_{jk}A_{ki} + \frac{1}{3!} x_jA_{jk}A_{kl}A_{li} = x_j \left( e^A \right)_{ji}
$$
or, better for our purpose here,
$$
e^{ix_jA_{jk}p_k} x_i e^{-ix_jA_{jk}p_k} = x_i + x_j (e^A - I)_{ji}
$$
Multiply on the left by $e^{ix_jA_{jk}p_k}$, rearrange, and obtain
$$
\left[x_i, e^{ix_jA_{jk}p_k} \right] = x_j \left( I - e^A \right)_{ji} e^{-ix_l A_{lk} p_k}
$$
For the commutator in the question title, $\left[x, e^{ipx}\right]$, the same procedure gives a much simpler formula:
$$
e^{ixp} x e^{-ixp} = x + [ixp, x] + \frac{1}{2!}[ixp,[ixp,x]] + \frac{1}{3!}[ixp,[ixp,[ixp,x ]]]\; + \;... =\\
= x + x + \frac{1}{2!}x + \frac{1}{3!}x \;+\; ... = x + (e - 1)x
$$
and
$$
\left[x, e^{ixp} \right] = (1-e)xe^{ixp}
$$
