The domain is where $g(x)=\prod_{k=1}^8\sin(k\pi x)\gt0;$ $g$ is symmetric about $x=\tfrac12$

Count the zeros $\tfrac{k}{n}$ in $\big(0,\tfrac12\big)$ and note the sign fails to flip at $\tfrac14$ and $\tfrac13$

Let $g(x)=\prod_{k=1}^8\sin(k\pi x);$ the domain of $f$ is where $g(x)\gt0.$ Since $\sin(k\pi(1-x))$ $=(-1)^{k+1}\sin(k\pi x)$ and $\sum_{k=1}^8(k+1)$ is even, $g(1-x)=g(x),$ so it suffices to study $\big(0,\tfrac12\big)$ and double.

In $\big(0,\tfrac12\big)$ the zeros of $g$ are the fractions $\tfrac{k}{n}$ with $2\le n\le8,$ $1\le k\lt\tfrac n2,$ and $\gcd(k,n)=1.$ For $n=2,\ldots,8$ there are $0,1,1,2,1,3,2$ of them, totaling $10.$

These $10$ zeros split $\big(0,\tfrac12\big)$ into $11$ subintervals on which $g$ has constant sign. Near $0$ every factor is positive, so $g\gt0$ there, and the sign flips at each zero except $\tfrac14=\tfrac28$ and $\tfrac13=\tfrac26,$ where an even number of factors vanish.

Tracking the signs, exactly $6$ of the $11$ subintervals have $g\gt0.$ By symmetry there are $6$ more in $\big(\tfrac12,1\big),$ so $n=12.$

Thus, B is the correct answer.