[This answer is still a work in progress. I'll get back to it later]. You can calculate an expression for this, for any n.
Let your volume be V.
By scaling, the volume of the set {|A|≤K} is VK<sup>n<sup>2</sup></sup>. Now let M be a matrix whose entries are independent normal random variables with mean 0 variance 1. From the density function of the normal distribution, this gives P(|M|≤K)~(2π)<sup>-n<sup>2</sup>/2</sup>VK<sup>n<sup>2</sup></sup> in the limit of small K.
I'll now calculate this expression in an alternative way. Use the M=QR decomposition, where Q is orthogonal and R is upper triangular, with diagonal elements λ<sub>n</sub>, λ<sub>n-1</sub>,…λ<sub>1</sub>. This can be done in such a way that λ<sub>k</sub><sup>2</sup> has the <a href="http://en.wikipedia.org/wiki/Chi%5E2">χ<sup>2</sup><sub>k</sub>-distribution</a> (a quick google search gives <a href="http://www.nowpublishers.com/product.aspx?product=CIT&doi=0100000001§ion=x1-36r1">this</a> but there's probably better references). The upper triangular parts of R have the standard normal density. I have to think how we can calculate the |R|. I was originally thinking that this is the max eigenvalue, but it's not quite that simple.