I think there is a construction. I will explain in 3D. $ABCD$ is the tetrahedron. Take the two opposite edges $AB$ and $CD$ and their midpoints $M_{AB}$ and $M_{CD}$ and connect them with a segment, denoted by $l$. Now take two planes $\alpha_1$ and $\alpha_2$, each parallel to both $AB$ and $CD$, intersecting the tetrahedron. Then, the intersection of $\alpha_i$ and ABCD is a parallelogram $p_i$ for $i=1,2$. Then $l$ passes through the centers (intersection point of the diagonals, centroids, ets...) $O_1$ and $O_2$ of $p_1$ and $p_2$ respectively. Translate $p_2$ on $\alpha_1$ along the vector determined by $O_2O_1$ and $p_1$ on $\alpha_2$ along $O_1O_2$. The intersections $p_1$ and $p_2 + O_2O_1$, and $p_2$ and $p_1 + O_1O_2$ gives you a parallelepiped. This is the one you need. Now, by varying the planes $\alpha_1$ and $\alpha_2$ as well as the choices of pairs of opposite edges, it seems to me, you get the family you need.
I hope I am not too far off, but maybe you can get the idea.