First, we know $ \sin 60^\circ = \sin 120^\circ = \frac{\sqrt{3}}{2}, $ so \begin{align*} &\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ \sin 100^\circ \sin 120^\circ \sin 140^\circ \sin 160^\circ \\ &= \frac{3}{4} \sin 20^\circ \sin 40^\circ \sin 80^\circ \sin 100^\circ \sin 140^\circ \sin 160^\circ.\end{align*}We can then write $ \sin 80^\circ = \sin 100^\circ = \cos 10^\circ, $ $ \sin 140^\circ = \sin 40^\circ, $ $ \sin 160^\circ = \sin 20^\circ, $ so \begin{align*} \frac{3}{4} \sin 20^\circ \sin 40^\circ \sin 80^\circ \sin 100^\circ \sin 140^\circ \sin 160^\circ &= \frac{3}{4} \cos^2 10^\circ \sin^2 20^\circ \sin^2 40^\circ \\ &= \frac{3}{4} (\cos 10^\circ \sin 20^\circ \sin 40^\circ)^2.\end{align*}By product-to-sum, \begin{align*} \cos 10^\circ \sin 20^\circ \sin 40^\circ &= \cos 10^\circ \cdot \frac{1}{2} (\cos 20^\circ - \cos 60^\circ) \\ &= \frac{1}{2} \cos 10^\circ \left( \cos 20^\circ - \frac{1}{2} \right) \\ &= \frac{1}{2} \cos 10^\circ \cos 20^\circ - \frac{1}{4} \cos 10^\circ \\ &= \frac{1}{4} (\cos 30^\circ + \cos 10^\circ) - \frac{1}{4} \cos 10^\circ \\ &= \frac{1}{4} \cos 30^\circ \\ &= \frac{\sqrt{3}}{8}.\end{align*}Therefore, the expression is equal to $ \frac{3}{4} \left( \frac{\sqrt{3}}{8} \right)^2 = \boxed{\frac{9}{256}} $.