In general, the Zariski tangent space of $\textrm{Hilb}(X)$ at $[V]$ is naturally isomorphic to  $\textrm{Hom}_V(I_V/I_V^2, \, \mathcal{O}_V)$. 

When $X$ and $V$ are both smooth and projective, this group equals $H^0(V, \, N_{V/X})$. Therefore in your case we can write $$m \leq \dim _{[V]}\textrm{Hilb}(X) \leq \dim T_{[V]} \textrm{Hilb}(X) = H^0(V, \, N_{V/X}) \leq m.$$

This means $$\dim \textrm{Hilb}(X) = \dim T_{[V]} \textrm{Hilb}(X)=m,$$
that is, $\textrm{Hilb}(X)$ is smooth at $[V]$, of dimension $m$.