smooth equidimensional fibers over a smooth base

The flatness statement you would like follows from Theorem 3.3.27 of Schoutens's book on ultraproducts.

One can also show directly that $X$ is smooth (assuming it is irreducible or even just equidimensional). The problem is local on $X$ and $Y$ so we may assume that $X\subset \mathbb{C}^N$ is affine of codimension $k=N-n$ and $Y=\mathbb{C}^d$. Choose polynomials $f_1,...,f_m, t_1,...,t_d\in \mathbb{C}[X_1,...,X_N]$ such that $$ X\ = \ \{x\in\mathbb{C}^N\,:\, f_1(x)=\cdots = f_m(x)=0\} $$ and $\varphi(x)=(t_1(x),...,t_d(x))$ for all $x\in X$. Fix $p\in X$ and assume $0=f(p)\in\mathbb{C}^d$. Then

$$ X_0\ = \ \{x\in\mathbb{C}^N\,:\, f_1(x)=\cdots =f_m(x)=t_1(x)=\cdots =t_d(x)=0\ \} $$ Since $p$ is a smooth point of $X_0$, after reordering the indices if necessary, there exist $r\leq m$ and $s\leq d$ such that $r+s=k+d$ and the following set is linearly independent.

$$ \{df_1(p),...,df_r(p),dt_1(p),...,dt_s(p)\} $$ Since $s\leq d$ we see $r\geq k$. But $X$ has pure codimension $k$ so $r\leq k$. Thus $r=k$ and $\{df_1(p),...,df_k(p)\}$ is linearly independent so $p\in X$ is a smooth point.