Take $n=4$ and let $V = \{ z_1=z_2=0 \} \cup \{ z_3=z_4=0 \}$. I claim that $\mathbb{C}^4 \setminus V$ is homotopic to $S^3 \times S^3$, which has nontrivial homology in degree $6$, contrary to your supposed bound, which is in degree $5$.

Note that $\mathbb{C}^4 \setminus V = \left( \mathbb{C}^2 \setminus \{ (0,0) \} \right)^2$. Taking the quotient by $\mathbb{R}_{+}$, we see that $\mathbb{C}^2 \setminus \{ (0,0) \}$ is
homotopic to $S^3$, so $\mathbb{C}^4 \setminus V$ is homotopic to $S^3 \times S^3$.

I can prove the required cohomology vanishing if you require that $V$ be Cohen-Macaulay.

Write $U$ for $\mathbb{C}^4 \setminus V$. We have the Hodge-de Rham spectral sequence: $H^q(U, \Omega^p) \implies H^{p+q}(U, \mathbb{C})$. Singe $U$ is an open subset of $\mathbb{C}^n$, we have $\Omega^p \cong \mathcal{O}^{\oplus \binom{n}{p}}$ so $H^q(U, \Omega^p) \cong H^q(U, \mathcal{O})^{\bigoplus \binom{n}{p}}$.

We can identify $H^q(U, \mathcal{O})$ with a local cohomology module of $V$, which the Cohen-Macaulay condition should force to be $0$ for $q > n-k-1$. So $H^q(\Omega^p)$'s will be zero for $q>n-k-1$. Then the spectral sequence immediately forces cohomology to vanish for $p+q > n+(n-k-1)$, as you desired.

I have no idea of how to get a statement in homotopy out of the Cohen-Macaulay condition.