Let $X/k$ be a variety and $A/k$ be an abelian variety. Set $\mathcal{A} = A \times_k X$ and $\bar{X} = X \times_k \bar{k}$.
Is it true that $\mathcal{A}[\ell^n] = A[\ell^n] \times_k X$? And what is $\mathcal{A}[\ell^n](\bar{X})$?
Let $X/k$ be a variety and $A/k$ be an abelian variety. Set $\mathcal{A} = A \times_k X$ and $\bar{X} = X \times_k \bar{k}$.
Is it true that $\mathcal{A}[\ell^n] = A[\ell^n] \times_k X$? And what is $\mathcal{A}[\ell^n](\bar{X})$?
Assuming that by $\mathcal A[\ell^n]$ you mean the subgroup scheme of $\mathcal A$ consisting of the (scheme-theoretic) kernel of multiplication by $\ell^n$, then yes, it is true that $\mathcal A[\ell^n] = A[\ell^n]\times_k X$. This follows from the description of $\mathcal A[\ell^n]$ as the fibre product of the map $\ell^n:\mathcal A \to \mathcal A$ and the map $X \to \mathcal A^n$ giving the identity section (and the analogous description of $A[\ell^n]$).
Also (assuming that $\overline{X}$ is reduced or that $\ell$ is prime to the characteristic of $k$), $\mathcal A[\ell^n](\overline{X})$ will be a product of copies of $A[\ell^n](\overline{k})$, one copy for each connected component of $\overline{X}$.