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: 
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 $(-1)^{\frac{n^2 - n}{2}}$.
EDIT---
 I think (now that I have seen other answers) this argument also works. We want the determinant of linear transformation say $T$ such that $vec(A^{T})=Tvec(A)$ (the $vec$ operator which writes a matrix as a long vector column wise). It is easy to see that $vec(A^{T})$ can be obtained by rearranging the rows of the vector $vec(A^{T})$. Each one of the row swappings makes your determinant multplied by (-1). Now how many swappings you have to make is the question. You have to swap every element beneath the main diagonal with another from above the diagonal. So there are $c=n(n-1)/2$ swappings. So your determinant is $(-1)^c$. 

PREVIOUS ATTEMPT--(BEFORE I UNDERSTOOD THE QUESTION PROPERLY)
First of all, note that $det(AB)=det(A)det(B)$, now appy this recursively. You will get $det(A^{t})=(det(A)det(A^{t-1}))=(det(A))^t$