Yes, there is an $n\times n$ lemma, and even an $\mathbb N\times\mathbb N$ lemma. The spectral sequence argument that Reid gives works. Another elementary proof uses the salamander lemma, a result of George Bergman's that I blogged about at SBS. It's exactly the same as the proof of the $3\times 3$ lemma I wrote up there.
Here's a counterexample to the $\mathbb Z\times\mathbb Z$ lemma. If you read about the salamander lemma, you'll understand how I came up with it. Here Z denotes $\mathbb Z$ and all non-zero maps are the identity
0 -> 0 -> 0 -> 0 -> 0
| | | |
v v v v
0 -> 0 -> 0 -> Z -> 0
| | | |
v v v v
0 -> 0 -> Z -> Z -> 0
| | | |
v v v v
0 -> Z -> Z -> 0 -> 0
.
.
.
Extend the diagram by copies of Z down and to the left, and put 0's everywhere else. All columns are exact, and all rows except one (the one with a single Z in it) are exact.