Can we find lattice polyhedra with faces of area 1,2,3,...?

I found a 32-face example with face areas $\{ 1, 2, \dots, 32 \}$:

It took a reasonable amount of experimentation to stop it from self-intersecting.

By using the Euler characteristic $\chi = L-E+V$ of the graph $\Gamma$ corresponding to the polyhedron (each cube is a vertex and we link adjacent cubes). One can show that $$ S = 4N - 2L + 2\chi $$ where $S$ is the total area of the uncovered faces, $N$ the number of cubes and $L$ the number of loops in $\Gamma$.

This gives a constraint on the possible $A(P)$. For example $A(P)=1^n 2^n 3^n \dots p^n$ can only be constructed if $4\mid np(p+1)$.

For the interesting case of golyhedra ($n=1$), our constraint reduces to $p \equiv 0,3 \:\mathrm{mod}\: 4$.

$p$-face golyhedra have been constructed by Alexey and Adam for $p=12, 15, 32$. Our constraint suggests that $p=11$ should also be possible. And indeed, we have found an 11-face golyhedron (see image). Because $p=7,8$ are easily ruled out, this has to be the smallest golyhedron.

We are led to conjecture that a $p$-face golyhedron exists if and only if $p \equiv 0,3 \:\mathrm{mod}\: 4$.