I don't know if this is the optimal, but an isosceles triangle with base and height $\sqrt{2}$ overlaps $2 \left(\sqrt{2}-1\right) \approx 0.828427$ when placed as below, and so improves over $\frac{3}{4}$:
[![SquareTri][1]][1]
Added. A tetrahedron formed from the above triangle and a point over the center of the cube. Just an image—no computations, no claims:
[![CubeTetra][2]][2]
Tetrahedron with same unit-area isosceles base, and height $3$.