Let \[g(x) = \sin (\pi x) \cdot \sin (2 \pi x) \cdot \sin (3 \pi x) \dotsm \sin (8 \pi x).\]Then the domain of $ f(x) $ is the set of all $ x $ such that $ g(x) > 0 $. The points where $ g(x) = 0 $ are the points of the form $ x = \frac{k}{n}, $ where $ 1 \le n \le 8 $ and $ 0 \le k \le n $. Since \[\sin (n \pi (1 - x)) = (-1)^{n + 1} \sin (n \pi x),\]we have that $ g(1 - x) = g(x) $. Also, $ g \left( \frac{1}{2} \right) = 0, $ so it suffices to consider the points where $ x \le \frac{1}{2} $. These points, increasing order, are \[x_0 = 0, \ x_1 = \frac{1}{8}, \ x_2 = \frac{1}{7}, \ x_3 = \frac{1}{6}, \ x_4 = \frac{1}{5}, \ x_5 = \frac{1}{4}, \ x_6 = \frac{2}{7}, \ x_7 = \frac{1}{3}, \ x_8 = \frac{3}{8}, \ x_9 = \frac{2}{5}, \ x_{10} = \frac{3}{7}, \ x_{11} = \frac{1}{2}.\]As $ x $ increases from 0 to $ \frac{1}{2}, $ as $ x $ passes through each point $ x_i, $ a number of the factors of the form $ \sin (n \pi x) $ will change sign. We list the $ n $-values for each value of $ i $: \[ \begin{array}{c|c} i & n \\ \hline 1 & 8 \\ 2 & 7 \\ 3 & 6 \\ 4 & 5 \\ 5 & 4, 8 \\ 6 & 7 \\ 7 & 3, 6 \\ 8 & 8 \\ 9 & 5 \\ 10 & 7 \\ 11 & 2, 4, 6, 8 \end{array} \]For example, as $ x $ increases, from being just less than $ x_1 = \frac{1}{8} $ to just greater than $ x_1, $ only $ \sin (8 \pi x) $ changes sign, from positive to negative. Since $ f(x) $ is positive on the interval $ (0,x_1), $ it will be negative on the interval $ (x_1,x_2), $ and so on. Thus, we can compute the sign of $ f(x) $ on each interval: \[ \begin{array}{c|c} i & \text{Sign of $ g(x) $ on $ (x_i,x_{i + 1}) $} \\ \hline 0 & + \\ 1 & - \\ 2 & + \\ 3 & - \\ 4 & + \\ 5 & + \\ 6 & - \\ 7 & - \\ 8 & + \\ 9 & - \\ 10 & + \\ 11 & - \end{array} \]We see that $ f(x) $ is positive on 6 intervals less than $ \frac{1}{2}, $ so $ f(x) $ is positive on 6 intervals greater than $ \frac{1}{2} $. This gives us a total of $ \boxed{12} $ intervals.