5. Let the right triangular prism $A B C-A_{1} B_{1} C_{1}$ have side edges and base edges of length 1. Then the common volume of the tetrahedra $A_{1} A B C$, $B_{1} A B C$, and $C_{1} A B C$ is $\qquad$
5. $\frac{\sqrt{3}}{36}$.

By symmetry, the intersection point $O$ of plane $C_{1} A B$, plane $A_{1} B C$, and plane $B_{1} A C$ projects onto the center $H$ of the equilateral $\triangle A B C$ on the base $A B C$.

As shown in Figure 2, let the midpoints of $A B$ and $B C$ be $D$ and $E$, respectively.

Then $O H$ is the intersection line of $\triangle E A A_{1}$ and $\triangle D C C_{1}$
$$
\begin{aligned}
\Rightarrow & \frac{O H}{C_{1} C}=\frac{D H}{D C} \\
& =\frac{1}{3} \\
\Rightarrow & O H=\frac{1}{3} .
\end{aligned}
$$

Thus, the common part of tetrahedron $A_{1} A B C$, tetrahedron $B_{1} A B C$, and tetrahedron $C_{1} A B C$ is tetrahedron $O A B C$.
$$
\begin{array}{l}
\text { Therefore, } V_{\text {tetrahedron }}=\frac{1}{3} O H \cdot S_{\triangle A B C} \\
=\frac{1}{3} \times \frac{1}{3} \times \frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{36} .
\end{array}
$$