The following method reduces the problem to the known case $p=1$.
Notice that $P_Z \equiv ZZ^t$ is an orthogonal projector of rank $p$, thus:
$\det(Z^t A Z) = \det( I_p + Z^t(A-I_m)Z) = \det( I_m + ZZ^t(A-I_m)ZZ^t) = \det( I_m + P_Z(A-I_m)P_Z)$
Where, $I_p$, $I_m$ are the $p$ and $m$ dimensional unit matrices. The second equality is due to the fact that the matrices $Z^t(A-I_m)Z$, $ZZ^t(A-I_m)ZZ^t$ have the same secular coefficients, by cyclic permutations and the identity $Z^tZ = I_p$.
Denote by $\rho(A)$ the completely antisymmetric $p$-tensor product representation of $A$:
$\rho(A) \circ v_1 \wedge . . . \wedge v_p = Av_1 \wedge . . . \wedge A v_p$
We have:
$\det( I_m + P_Z(A-I_m)P_Z) = \mbox{tr} (\rho(P_Z) \rho(A))$
(Both sides are equal, because they are just the determinant of the restriction of A to the range of $P_Z$.
Now,$\rho(P_Z)$ is a one dimensional projector over the $p$-wedge product of the range of $P_Z$, and the eigenvalues of $\rho(A)$ are just all possible products of $p$ distinct eigenvalues. By the known rank one solution, the absolute minimum is the minimal eigenvalues of $\rho(A)$ which is the product of the $p$ smallest eigenvalues of $A$.