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.

Edit 1: Indeed, my previous suggestion was not good. As it has been pointed out to me by Dongryul Kim, the parallelotopes should be finitely many. I have overlooked that. Maybe this could work?

Construction 1:

In the case of a triangle $ABC$ we define a parallelogram at vertex $A$ to be a parallelogram constructed as follows: pick a point $Q$ on $BC$ and draw the two lines parallel to $AB$ and $AC$.

Induction: Take a $d-1$ face $f$ of the $d$-simplex and the opposite vertex $v$ (the one vertex which is not on $f$). Take a vertex $w$ of $f$. Look at the edge $vw$. Take a point $p$ on $vw$ and draw the hyper-plane $L$ through $p$ parallel to $f$. $L$ intersects the simplex in a $d-1$ simplex $f'$ similar to $f$ (actually homothetic from vertex $v$). Draw the parallelotope on $f'$ with vertex $p$ (see first step above or inductive step). Translate it to $f$ by the vector $\overrightarrow{pw}$ to obtain the desired $d$-dimensional parallelotope at vertex $w$.

Construction 2:

A parallelogram sitting on the edge $AB$ of triangle $ABC$ is a parallelogram defined as follows: Take midpoint $M$ on $AB$ and pick a point $Q$ on $CM$. Draw line parallel to $AB$ to form a segment parallel to $AB$ and then translate it down to $AB$ with vector $\overrightarrow{QM}$.

Induction: Pick a face $d-1$ dimensional face $f$ of the $d$-simplex and let $v$ be the vertex opposite to it (like before). Let $M$ be the barycenter of $f$ and choose a point $p$ on $vM$. Draw a $d-1$ hyperplane $L$ through $p$ parallel to $f$. $L$ intersects the $d$-simplex in a $d-1$ simplex $f'$ homothetic to $f$ (and parallel). choose a $d-2$ face $f''$ of $f'$ and draw the parallelotope sitting on $f''$ by induction. Now, translate it via the vector $\overrightarrow{pM}$ to $f$. Thus, one obtains a $d$-parallelotope sitting on $f$.

Seem like using these two procedures one can cover the simplex with finitely many parallelotopes for large enough $N$, by various combinations of face $f$, vertex $w \in f$ and point $p \in vw$ choices. Like various midpoints / barycenters for instance. Maybe only the first procedure is enough, but the second may lower the number of parallelotopes. I guess it depends on the problem. Do we need an estimate on the number $N$ in terms of $d$?