11.5 .

According to the equivalence, we only need to consider the cutting situation of a unit cube.

On the one hand, first, we need to show that 4 is not enough. If there were 4, since all faces of a tetrahedron are triangles and are not parallel to each other, the top face of the cube would have to be cut into at least two triangles, and the bottom face would also have to be cut into at least two triangles. The area of each triangle is less than or equal to $\frac{1}{2}$, and these four triangles must belong to four different tetrahedrons. The height of a tetrahedron with such a triangle as its base is less than or equal to 1.

Therefore, the sum of the volumes of the four different tetrahedrons is less than or equal to $4\left(\frac{1}{3} \times \frac{1}{2} \times 1\right)=\frac{2}{3}<1$, which does not meet the requirement.
Thus, $k \geqslant 5$.
On the other hand, as shown in Figure 5, the unit cube can be cut into 5 tetrahedrons, for example, by removing a tetrahedron $A_{1} B C_{1} D$ from the center of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, leaving four tetrahedrons at the corners.

In total, there are 5 tetrahedrons.