To summarize what was indicated in the comments:
as @DerekHolt points out, $d(A_{k_1}^{e_1}\times\dots\times A_{k_r}^{e_r})=\max_i d(A_{k_i}^{e_i})$. This is a simple observation: if one has $d$ generators for $A_{k_i}^{e_i}$ for all $i$, this is a surjective homomorphism $\phi_i:F_d\to A_{k_i}^{e_i}$, where $F_d$ denotes the free group on $d$ generators. Then the diagonal $\phi_1\times \cdots\times \phi_r$ will have image generating $d(A_{k_1}^{e_1}\times\dots\times A_{k_r}^{e_r})$. To see this, note the subgroup it generates will surject each factor, and thus will have $A_{k_i}^{e_i}$ in its composition series. By uniqueness of the composition series (Jordan-Hölder Theorem), the group must therefore be a product of these factors.
As you note, Philip Hall proved that the maximal number $n$ such that $G^n$ is $k$-generated is $h_k(G)$, the Eulerian function, which counts the number of $k$-generating sets for $G$, divided by $|Aut(G)|$. Thus, the problem is reduced to determining the number of $k$-generating sets for $G$ for all $k$ (assuming one knows $Aut(G)$). In the literature, this is usually expressed as the probability that $k$ elements generate, so is divided by $|G|^k$ to get a number $< 1$, denoted by $p_k(G)$.
For $A_n$, we have $Aut(A_n)=S_n$ for $n\geq 7$. The best estimates on $p_2(A_n)$ appear to be
$$1−\frac{1}{n}−\frac{8.8}{n^2} ≤ p_2(A_n)<1−\frac{1}{n}−\frac{0.93}{n^2}.$$
Putting these facts together should give the maximal known possible value of $e_r$ that you desire.