等比数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和为 $S_{n}$, 若 $S_{2 n}=3\left(a_{1}+a_{3}+a_{5}+\cdots \cdots+a_{2 n-1}\right)\left(n \in \mathrm{N}^{*}\right)$, $a_{1} a_{2} a_{3}=8$, 则 $S_{8}=(\quad)$
设等比数列 $\left\{a_{n}\right\}$ 的公比为 $q$,

$\because S_{2 n}=3\left(a_{1}+a_{3}+a_{5}+\cdots+a_{2 n-1}\right)\left(n \in \mathrm{N}^{*}\right)$,

$\frac{a_{1}}{1-q} \cdot\left(1-q^{2 n}\right)=3 \cdot \frac{a_{1}}{1-q^{2}} \cdot\left(1-q^{2 n}\right)$,

$\therefore q=2$,

$\because a_{1} a_{2} a_{3}=a_{2}^{3}=8$,

$\therefore a_{2}=2, a_{1}=1$,

则 $S_{8}=\frac{1-2^{8}}{1-2}=255$.

故选: B.