Consider the matrices with a single 1 on the diagonal, 0 elsewhere. There are $n$ of these. There are also $\frac{n(n-1)}{2}$ matrices with two 1s in symmetric positions off diagonal and $\frac{n(n-1)}{2}$ with a 1 and -1 in symmetric positions off diagonal (so the resultant matrix is antisymmetric).
These form a basis for $M_n(\mathbb{R})$ (they are clearly linearly independent, and there are $n^2$ of them). The symmetric matrices we have defined has eigenvalue 1 under $\varphi$, and the antisymmetric matrices we have defined has eigenvalue -1. It should be obvious what the matrix of $\varphi$ looks like with respect to this basis: $n+\frac{n+1}{2}$ 1s and $\frac{n(n-1)}{2}$ -1s down the diagonal, zero elsewhere.
The determinant of $\varphi$ is the product of eigenvalues: $(-1)^{\frac{n(n-1)}{2}}$