8. Given a regular tetrahedron $A_{1} A_{2} A_{3} A_{4}$, four parallel planes $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}$ are drawn through its four vertices. If the distance between each pair of adjacent planes is 1, then the volume of the tetrahedron is $\qquad$.
8. $\frac{5 \sqrt{5}}{3}$.

As shown in Figure 3, a regular tetrahedron is complemented to form a cube, $E_{1}$ and $F_{1}$ are the midpoints of $A_{1} B_{1}$ and $C_{1} D_{1}$, respectively. The planes $E E_{1} D_{1} D$ and $B B_{1} F_{1} F$ are two parallel planes, and their distance is 1.
As shown in Figure 4, let the edge length of the cube be $a, A_{1} M=M N=1$. Then $A_{1} E_{1}=\frac{a}{2}, D_{1} E_{1}=\sqrt{A_{1} D_{1}^{2}+A_{1} E_{1}^{2}}=\frac{\sqrt{5} a}{2}$. From $A_{1} D_{1} \cdot A_{1} E_{1}=A_{1} M \cdot D_{1} E_{1}$, we get $a=\sqrt{5}$.
Thus, $V_{\text {regular tetrahedron }, A, A, 1, A_{4}}=a^{3}-4 \times \frac{1}{6} a^{3}=\frac{5 \sqrt{5}}{3}$.