8. If the positive real numbers $a_{1}, a_{2}, a_{3}, a_{4}$ satisfy $a_{1} \geqslant a_{2} a_{3}^{2}, a_{2} \geqslant a_{3} a_{4}^{2}, a_{3} \geqslant a_{4} a_{1}^{2}, a_{4} \geqslant a_{1} a_{2}^{2}$. Then the maximum value of $a_{1} a_{2} a_{3} a_{4}\left(a_{1}-a_{2} a_{3}^{2}\right)\left(a_{2}-a_{3} a_{4}^{2}\right)\left(a_{3}-a_{4} a_{1}^{2}\right)\left(a_{4}-a_{1} a_{2}^{2}\right)$ is
$$
\begin{array}{l}
a_{1} a_{2} a_{3} a_{4}\left(a_{1}-a_{2} a_{3}^{2}\right)\left(a_{2}-a_{3} a_{4}^{2}\right)\left(a_{3}-a_{4} a_{1}^{2}\right)\left(a_{4}-a_{1} a_{2}^{2}\right) \\
=a_{2}\left(a_{1}-a_{2} a_{3}^{2}\right) \cdot a_{3}\left(a_{2}-a_{3} a_{4}^{2}\right) \cdot a_{4}\left(a_{3}-a_{4} a_{1}^{2}\right) \cdot a_{1}\left(a_{4}-a_{1} a_{2}^{2}\right) \\
=\left(a_{1} a_{2}-a_{2}^{2} a_{3}^{2}\right)\left(a_{2} a_{3}-a_{3}^{2} a_{4}^{2}\right)\left(a_{3} a_{4}-a_{4}^{2} a_{1}^{2}\right)\left(a_{4} a_{1}-a_{1}^{2} a_{2}^{2}\right) \\
\leqslant\left[\frac{\left(a_{1} a_{2}-a_{1}^{2} a_{2}^{2}\right)+\left(a_{2} a_{3}-a_{2}^{2} a_{3}^{2}\right)+\left(a_{3} a_{4}-a_{3}^{2} a_{4}^{2}\right)+\left(a_{4} a_{1}-a_{4}^{2} a_{1}^{2}\right)}{4}\right]^{4} .
\end{array}
$$

Notice that $a_{1} a_{2}-a_{1}^{2} a_{2}^{2}=a_{1} a_{2}\left(1-a_{1} a_{2}\right) \leqslant\left(\frac{a_{1} a_{2}+1-a_{1} a_{2}}{2}\right)^{2}=\frac{1}{4}$,
Similarly, we get $a_{2} a_{3}-a_{2}^{2} a_{3}^{2} \leqslant \frac{1}{4}, a_{3} a_{4}-a_{3}^{2} a_{4}^{2} \leqslant \frac{1}{4}, a_{4} a_{1}-a_{4}^{2} a_{1}^{2} \leqslant \frac{1}{4}$,
Therefore, $a_{1} a_{2} a_{3} a_{4}\left(a_{1}-a_{2} a_{3}^{2}\right)\left(a_{2}-a_{3} a_{4}^{2}\right)\left(a_{3}-a_{4} a_{1}^{2}\right)\left(a_{4}-a_{1} a_{2}^{2}\right) \leqslant \frac{1}{4^{4}}=\frac{1}{256}$, equality holds when $a_{1}=a_{2}=a_{3}=a_{4}=\frac{\sqrt{2}}{2}$.