Solution:

The two regular tetrahedra use the two alternating sets of four vertices of the cube. Each has edge length $\sqrt2$, a face diagonal of the cube.

The volume of a regular tetrahedron with edge length $s$ is $\dfrac{\sqrt2}{12}s^3$. Thus one large tetrahedron has volume $\dfrac{\sqrt2}{12}(\sqrt2)^3=\dfrac13$.

Intersect one tetrahedron with the other. Each face of the first cuts from the second a corner tetrahedron similar to the original with scale factor $\dfrac12$, so each cut-off piece has $\dfrac18$ of the large tetrahedron's volume.

There are four such corner pieces, so the intersection has $1-4\cdot\dfrac18=\dfrac12$ of the volume of one large tetrahedron. Hence the intersection volume is $\dfrac12\cdot\dfrac13=\dfrac16$.

Thus, D is the correct answer.