[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 VKⁿ². 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)^-n2/2VKⁿ² 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_n, λ_n-1,…λ₁. This can be done in such a way that λ_k² has the χ²_k-distribution (a quick google search gives this 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.

I had a go at this question, but the method I tried here doesn't quite work out. It does reduce it to upper triangular matrices, although that doesn't seem to be a lot of help for general n.

Let your volume be V.

By scaling, the volume of the set {|A|≤K} is VKⁿ². 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)^-n2/2VKⁿ² 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_n, λ_n-1,…λ₁, which are the eigenvalues of R. This can be done in such a way that λ_k² has the χ²_k-distribution (a quick google search gives this but there's probably better references). The upper triangular parts of R have the standard normal density. We need to calculate |R|. I was originally thinking that this is the max eigenvalue, but it's not quite that simple.

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 VKⁿ². 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)^-n2/2VKⁿ² 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_n, λ_n-1,…λ₁. This can be done in such a way that 2nλ_k² has the χ²_2k-distribution (a quick google search gives this but there's probably better references).

P(|M|≤K)=P(|R|≤K)=P(max λ_k²≤K²)=∏_kP(2nλ_k²≤2nK²).

The idea is now to put in the cumulative density functions and solve for V from the previous expression. Let me check through the details first...

[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 VKⁿ². 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)^-n2/2VKⁿ² 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_n, λ_n-1,…λ₁. This can be done in such a way that λ_k² has the χ²_k-distribution (a quick google search gives this 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.