If the intersection $\cap_{i=1}^nV_i$ is irreducible, then it is equidimensional. Otherwise let $r<n$ be such that $\cap_{i=1}^rV_i$ is irreducible, but $A:=\cap_{i=1}^{r+1}V_i$ is not. If $\dim\cap_{i=1}^rV_i\geq 2$, then $A$, being an effective ample divisor, is connected, so if it is smooth, then it is irreducible, contradicting the choice of $r$, so $\dim\cap_{i=1}^rV_i\leq 1$. Then $A$ is either empty or a finite set and hence equidimensional. (You can also remark, that it is not needed that all intersections are smooth, just that there is a sequence of getting to $\dim\cap_{i=1}^nV_i$ through smooth intersections).

On the other hand, if you only assumed that the ultimate intersection is smooth, then this is not true:
Let $V_1,V_2\subseteq \mathbb P^3$ be two quadric surfaces that share a tangent plane. Then $V_1\cap V_2$ is the union of two intersecting lines, say $\ell_1, \ell_2$. Now let $V_3$ be a third quadric that contains $\ell_1$ but does not contain $\ell_2$. Then $V_3$ intersects $\ell_2$ in two distinct points, one of which is on $\ell_1$. Let $P$ be the intersection point of $V_3$ and $\ell_2$ which is not on $\ell_1$. Then
$V_1\cap V_2\cap V_3=\ell_1\cup \{P\}$. Which is smooth but not equidimensional. Of course, this can only be done if the intersection is the union of a positive dimensional irreducible component and a set of points.