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][1]. 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. All non-zero maps are the identity

$$\require{AMScd}
\begin{CD}
0 @>>> 0 @>>> 0 @>>> 0 @>>> 0\\
@.    @VVV    @VVV   @VVV   @VVV\\
0 @>>> 0 @>>> 0 @>>> \mathbb{Z} @>>> 0\\
@.     @VVV   @VVV   @VVV            @VVV\\
0 @>>> 0 @>>> \mathbb{Z} @>>> \mathbb{Z} @>>> 0 \\
@.     @VVV   @VVV             @VVV           @VVV\\
0 @>>> \mathbb{Z} @>>> \mathbb{Z} @>>> 0 @>>> 0
\end{CD}
$$

Extend the diagram by copies of $\mathbb Z$ 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 $\mathbb Z$ in it) are exact.



  [1]: http://sbseminar.wordpress.com/2007/11/13/anton-geraschenko-the-salamander-lemma/