Let $H:\{0,1\}^*\to\{0,1\}^n$ be a cryptographic hash function as a black-box, and suppose we have unlimited space.
As I understand, finding $x$ such that $H(x)=0$ (if such exists) would require a preimage attack, and avg. time $O(2^n)$ (linear in the size of the output). On the other hand, finding $x\neq y$ such that $H(x)\oplus H(y)=0$ could use the birthday attack, and therefore avg. time $O(2^{n/2})$.
My question is, if something better can be said for finding 3 (or more) distinct values $x,y,z$ such that $H(x)\oplus H(y)\oplus H(z)=0$. It seems clear that this could be done using less samples of $H$, but it's not clear to me if the time complexity could be improved.