For clarity, the best way to work with this is complex-analytically. I am sure there is a, probably more involved, algebraic proof.

Lemma: Let $M$ be a complex manifold and let $X$ be a $\mathbb C^\times$-bundle on $X$. Let $Y$ be a finite etale cover of $X$. Then $Y$ is a $\mathbb C^\times$ bundle on an etale cover of $M$, with that bundle being an $n$th tensor root of the pullback of $X$.

Since etale covers of abelian varieties are just isogenies, that gives you the explicit description.

Proof of the lemma: Consider the inverse image in $Y$ of a fiber of $X$ over $M$. This is a union of connected components. The components, being etale covers of $\mathbb C^\times$, are copies of $\mathbb C^\times$ that map to it along an $n$th power map. Let $N$ be $Y$ with each connected component contracted. That is, it is the quotient by the equivalence relation that two points are equivalent if they are in the same connected component of a fiber over $M$. Then $Y$ is a $\mathbb C^\times$-bundle on $N$.

$N$ has a map to $M$. We prove that it is etale. This is local on $M$, so consider an open ball on which $X$ is trivial. Then $X$ is just $\mathbb C^\times$ cross an open ball. The fundamental group is $\mathbb Z$, so all etale covers are just the obvious $n$th power maps, and in all of these the map $N\to M$ is etale.

Furthermore these obvious $n$th power maps are locally $n$th power maps, and $n$ is the same in the entire open ball, therefore locally constant, therefore constant. So the map from the $Y$ bundle to the pullback of $X$ is an $n$th power map, so the pullback of $X$ is the $n$th tensor power.