Let $X$ be an abelian variety over a field $k$.

Let $A^p_{\rm hom}(X)$ be the $p$-th Chow group of cycles modulo homological equivalence ($\ell$-adic, if $k$ is of char $p$).

Do we have $$A^p_{\rm hom}(X) = \wedge^p A^1_{\rm hom}(X)?$$

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No. Because Chow groups lie in the even-dimensional homology, their multiplication structure is commutative. So the natural map is from $\operatorname{Sym}^p A^1$ to $A^p$.

Also no for dimensions reasons. For instance $p=2$, $A$ is a product of two isogenous elliptic curves, $A^1$ is three-dimensional (four if they're CM) but $A^2$ is one-dimensional.