Question: So if I define a linear transformation $ T: M\_{n\times n}(R) \rightarrow M\_{n\times n}(R) $ and $ T(A)=A^t $ what would be its determinant?

Answer: 
The determinant is most easily determined in a basis of eigenvectors. There are $n$ matrices with a single $1$ on the diagonal, $n(n-1)/2$ matrices with two $1$s in symmetric positions and $n(n-1)/2$ matrices with a $1$ and a $-1$ in symmetric positions. These form a basis together. The diagonal and symmetric matrices have eigenvalue $1$ under transposition, the antisymmetric matrices have eigenvalue $-1$, so the determinant is $(-1)^{n(n-1)/2}$, which is $+1$ if $n\bmod4$ is $0$ or $1$ and $-1$ if it's $2$ or $3$.
Let us construct the matrix representation with respect to the standard basis vectors ordered in a special way. The first $n$ are $\mathbb{e}\_{ii}$ for $1\le i \le n$. Then pair the remaining matrices in transpose pairs. If you construct the resulting matrix, you will have a block diagonal matrix beginning with an identity block along with $\frac{n^2 - n}{2}$ anti-diagonal blocks of size $2$. It follows that the determinant is given by $(-1)^{\frac{n^2 - n}{2}}$.