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 $A$.