The answer is yes. Let $I = q_1 \cap q_2 \cap \cdots \cap q_k$ be the primary decomposition of $I$. Let $p_i$ be the radical $\sqrt{q_i}$ and let $N_i$ be large enough that $q_i \supseteq p_i^{N_i}$. I claim that we can take $N= \max(N_i)$. We'll abbreviate $\mathbb{C}[x_1, \ldots, x_n]$ to $A$.

Let $f$ be a polynomial as in the statement of the question. In order to show that $f \in I$, it is enough to show that $f \in q_i$ for every $i$. We focus on one $q$ to pay attention to, so we can drop the subscript $i$ and just talk about $p$, $q$ and $N$. Let $W$ be the variety of $p$.

Now, $A$ is regular, so $A_p$ is regular. In other words, if $p$ has codimension $d$, there are transverse smooth hypersurfaces $y_1$, $y_2$, ..., $y_d$ which generate the maximal ideal of $A_p$. So there is some $u \not \in p$ such that $u^{-1} p = \langle y_1, \ldots, y_d \rangle$ in the localization $A[u^{-1}]$.

We claim that $f$, as an element of $A[u^{-1}]$ is in $u^{-1} p^N$. Proof: Write
$$f=\sum_{i_1+\cdots+i_d < N} b_{i_1 \ldots i_d} y_1^{i_1} \cdots y_d^{i_d} + r$$
with $r \in u^{-1} p^N$ and with all the $b$'s either $0$ or not in $p$. We want to show the $b$'s are zero. If not, let $(i_1, \ldots, i_d)$ be minimal with the corresponding $b$ nonzero. Since $b$ is not in $p$, it is not zero on $W$. Choose $w$ in $W$ where neither $b$ nor $u$ vanishes.

Since the $y_i$'s are transverse hypersurfaces, we can take $\sum i_j$ derivatives to obtain a polynomial with leading term $\mbox{nonzero stuff} \cdot \prod (i_j)! \cdot b$ at $w$. Since we are in characteristic zero, this doesn't vanish at $w$. (In finite characteristic, the factorials might be zero.) But $w \in W \subseteq V$, so this derivative is supposed to vanish at $w$. This contradiction shows that $f \in u^{-1} p^N$.

Since $q \supseteq p^N$, we have $f \in u^{-1} q$ so $u^k f \in q$ for some $k$. But $q$ is primary and $u \not \in p = \sqrt{q}$. So this shows $f \in q$, as desired.