Let $f:X\to \operatorname{Spec}\mathbb Z[1/n]$ be a morphism of finite type. As $f$ is generically flat, there is a finite set of primes $S$ such that $X\to \operatorname{Spec} \mathbb Z$ is flat over $\operatorname{Spec}\mathbb Z - S$. (This is easy to show, so let's assume $f$ is flat now.)
If $f$ is flat along the generic point, and its generic fiber is smooth, then it is smooth along the preimages of the generic point by a classical result. We can find an affine open neighborhood of each preimage of the generic point such that the restriction of $f$ to each affine open is smooth. Since $f$ is quasi-compact there are finitely many of these, call them $U_1,\ldots,U_n$. Taking the intersection $V = \cap_{i=1}^n f(U_i)$, we have an open set of $\operatorname{Spec}\mathbb{Z}$ for which the induced map $f: f^{-1}(V) \to V$ is smooth. The topology on $\operatorname{Spec}\mathbb{Z}$ is such that non-empty opens are the complement of finitely many primes, so there is a finite set of primes $S$ such that $X\to \operatorname{Spec}\mathbb Z[1/n]$ is smooth over $\operatorname{Spec}\mathbb Z - S$.
In general you can't weaken finite type to locally of finite type here.
To prove this, you can proceed in many ways. One way is to use the sheaf of differentials. That sheaf is locally free [Edit: of rank $\dim X_{\mathbb{Q}}$] if and only if the morphism is smooth.
