12.278 ** Let the volume of the regular tetrahedron $ABCD$ be $1$, and $O$ be its center. The regular tetrahedron $A'B'C'D'$ is symmetric to the regular tetrahedron $ABCD$ with respect to point $O$. Then the volume of the common part of these two regular tetrahedra is $\qquad$
If the tetrahedron $A-BCD$ is placed on a horizontal plane, it is easy to see that the distance from its center to point $A$ is $\frac{3}{4}$ of the distance from point $A$ to the base. Therefore, the reflective symmetry plane is a horizontal plane equidistant from point $A$ and the base. Thus, the small regular tetrahedron cut at point $A$ is a $\frac{1}{2}$ reduction of the original regular tetrahedron. Similarly, the regular tetrahedra cut at points $B$, $C$, and $D$ are also $\frac{1}{2}$ reductions of the original regular tetrahedron. After cutting out these four smaller regular tetrahedra from the original tetrahedron, the common part of the two regular tetrahedra is obtained, with a volume of $1-4 \times\left(\frac{1}{2}\right)^{3}=\frac{1}{2}$.