Yes. Morphism $\pi$ is smooth if $X$ is smooth. Smoothness is stable under base change, so $\pi$ is smooth implies $\pi^{-1}(Y) \to Y$ is smooth. By Lemma 35.17.4 in "Stack Project" "being regular" (which is equivalent to "being smooth" when the base field is $\mathbb C$) is a property local in the smooth topology, hence, $\pi^{-1}(Y)$ smooth, $\pi^{-1}(Y) \to Y$ smooth and surjective implies $Y$ is smooth.
Why $\pi$ is smooth if $X$ is smooth. $X$ is smooth iff defining fan $\Sigma$ is smooth. Consider fan $\tilde{\Sigma}$ in $\mathbb R^{\Sigma(1)}$ corresponding to toric variety $\mathbb C^{\Sigma(1)} - Z(\Sigma)$ which for every cone $\sigma \in \Sigma$ have cone $\tilde{\sigma} = \operatorname{cone}(e_{\rho} | \rho \in \sigma(1)) \in \tilde{\Sigma}$ and all corresponding faces (see Cox Little Schenck "Toric varieties" §5.1 for more accurate construction of $\tilde{\Sigma}$). We have to show that morphism $\pi : X_{\tilde{\Sigma}} \to X_{\Sigma}$ induced by canonical morphism of fans $\tilde{\Sigma} \to \Sigma$ is smooth. Since smoothness is a property local on source, it is enough to check smoothness of restriction $\pi|_{U_{\tilde{\sigma}}}$ for every affine open set $U_{\tilde{\sigma}}, \sigma \in \Sigma$. Computation in coordinates shows that morphism $\pi|_{U_{\tilde{\sigma}}} : U_{\tilde{\sigma}} \to U_{\sigma}$ is isomorphic to standard projection $\pi : \mathbb C^{k} \times (\mathbb C^*)^{\dim X-k} \times (\mathbb C^*)^{r} \to \mathbb C^{k} \times (\mathbb C^*)^{\dim X-k}$ where $k=\dim \sigma$, which is definitely smooth.