Let $V$ be a subvariety of $\mathbb C^n$ with irreducible components of dimension >$0$. Is $H_{2n1}(\mathbb C^n\setminus V)=0$?
It's certainly true, though I don't know the best way of proving it. One way to see it is that if you take the onepoint compactifications $V^+\subset(\mathbb{C}^n)^+=S^{2n}$, you get a connected space, since all components contain the point at infinity. The result then follows by Alexander duality in $S^{2n}$. 


The sharp bound is this: For any closed algebraic set $V$ of codimension $d$ in ${\Bbb C}^n$, with $U={\Bbb C}^n \setminus V$, one has $\pi_i(U) = 0$ for $0 < i\leq 2d2$ and $\pi_{2d1}(U) \neq 0$. Using the Hurewicz isomorphism, you get the same vanishing and nonvanishing for homology. A simple proof is in the appendix to my notes (with Fulton) on equivariant cohomology, http://www.math.washington.edu/~dandersn/eilenberg/ . A slick reason for vanishing was pointed out by David Speyer: given a (nice) map of an $i$sphere into $U$, the (real) lines between the points in image of the sphere and points in $V$ sweep out a space of dimension at most $(2n2d)+i+1$. When $i<2d1$, you can pick a point in $U$ not lying on any such line, and contract your sphere down to that point. The nonvanishing happens because all algebraic sets have nontrivial fundamental classes in BorelMoore homology. (Vanishing can also be proved using BM homology.) 

