jmc

We see that the given number is equal to $1\cdot n^3+1=n^3+1$. Using the sum of cubes factorization, it follows that $n^3+1=(n+1)(n^2-n+1)$. Since $1$ is a digit in the base, then $n>1$, and $n+1>1$ and $n^2-n+1>n-n+1=1$, so $n^3+1$ is the product of two integers greater than $1$. Thus, $1001_n$ is prime for $\boxed{0}$ values of $n$.