$\underline{\textbf{Solution 1}}$

Let the desired result be $S$ and multiply it with $\sin 20^\circ$: \begin{align} \sin 20^\circ\cdot S &= \sin 20^\circ \cos 20^\circ \cos 40^\circ \cos 80^\circ\\ &= \frac{1}{2}\cdot \sin 40^\circ \cos 40^\circ \cos 80^\circ \\ &= \frac{1}{4}\cdot \sin 80^\circ \cos 80^\circ \\ &= \frac{1}{8}\cdot \sin 160^\circ \\ &= \frac{1}{8}\cdot \sin 20^\circ \end{align} This means $$\sin 20^\circ\cdot S = \frac{1}{8}\cdot\sin 20^\circ \implies S=\boxed{\frac{1}{8}}$$

$\underline{\textbf{Solution 2}}$

This problem can also be solved by applying the triple angle formula. \begin{align} &\cos 20^\circ \cos 40^\circ \cos 80^\circ\\ &=\cos 20^\circ \cos (60^\circ - 20^\circ) \cos (60^\circ + 20^\circ)\\ &=\frac{1}{4}\cdot\cos(3\cdot 20^\circ)\\ &= \frac{1}{4}\cdot \frac{1}{4}\\ &= \boxed{\frac{1}{8}} \end{align}