Given an elliptic curve $E$, the Frobenius action $\mathrm{Frob}$, and $\mathrm{det}(1-\mathrm{Frob}_E T | H^1(E))$, how do we find an expression for:
$$\mathrm{det}(1-\mathrm{Frob}_X T | \wedge^2 (H^1(E) \otimes H^1(E)))$$
for $X=(E \times E)/(\mathbb{Z}/2\mathbb{Z})$, the abelian surface with 16 singular points which correspond to the 2-torsion points of $E$.
This seems to be an exercise in linear algebra but I am somewhat muddled with how operations work over exterior powers!
I am not sure at all if that's what you're looking for, but you're looking whether $\det{\Lambda^2f}$ can be determined as a function of $\det{f}$ when $f$ is an endomorphism of a finite-dimensional vector space?
It would be an algebraic relationship, so (assuming it exists) it would be enough to show it over the field of complex numbers for a dense subset of $f$, e.g. $f$ has $n$ distinct eigenvalues.
But when $f$ has $n$ distinct eigenvalues $\alpha_1,\ldots,\alpha_n$, with eigenvectors $e_i$, then the $e_i \wedge e_j$, $i < j$, are a basis of eigenvectors (with values $\alpha_i\alpha_j$) of $\Lambda^2f$. Therefore, $\det{\Lambda^2f} = \prod_{i < j}{\alpha_i\alpha_j}=\prod_{i=1}^n{\alpha_i^{n-i}}\prod_{j=1}^n{\alpha_j^{j-1}}=(\det{f})^{n-1}$.
So, in general, $\det{\Lambda^2f}=(\det{f})^{n-1}$, where $n$ is the dimension of the vector space $f$ is defined on.
Note that this would of course work for free modules over a commutative ring with unity.