One sees by inspection that $1$ and $-1$ are eigenvalues of $T$ (the vector $(1,0,0,0,0,0, -1)$ is an eigenvector for the eigenvalue $-1$).
Since, $T(e_i)=e_{8-i}$ the matrix representation of $T$ can be easily constructed: $ A=\begin{bmatrix}0&0&0&0&0&0&1\\ 0&0&0&0&0&1&0 \\0&0&0&0&1&0&0\\0&0&0&1&0&0&0 \\0&0&1&0&0&0&0\\0&1&0&0&0&0&0\\1&0&0&0&0&0&0 \end{bmatrix} $
For the eigenvalue $1$, we have $ A-I=\begin{bmatrix}-1&0&0&0&0&0&1\\ 0&-1&0&0&0&1&0 \\0&0&-1&0&1&0&0\\0&0&0&0&0&0&0 \\0&0&1&0&-1&0&0\\0&1&0&0&0&-1&0\\1&0&0&0&0&0&-1 \end{bmatrix} $ which has echelon form $ \begin{bmatrix}-1&0&0&0&0&0&1\\ 0&-1&0&0&0&1&0 \\0&0&-1&0&1&0&0\\0&0&0&0&0&0&0 \\0&0&0&0&0&0&0\\0&0&0&0&0&0&0\\0&0&0&0&0&0&0 \end{bmatrix} $ It can be deduced from the above that the eigenspace for the eigenvalue $1$ has dimension 4.
An echelon form of the matrix $T-(-1)I$ is $ \begin{bmatrix}1&0&0&0&0&0&1\\ 0&1&0&0&0&1&0 \\0&0&1&0&1&0&0\\0&0&0&2&0&0&0 \\0&0&0&0&0&0&0\\0&0&0&0&0&0&0\\0&0&0&0&0&0&0 \end{bmatrix} $ from which it follows that the dimension of the eigenspace for the eigenvalue $-1$ is 3.
Thus $T$ has exactly two eigenvalues $1$ and $-1$. Since the dimensions of the eigenspaces sum to 7, it follows that $T$ is diagonalizable. Since the product of the eigenvalues is $1^4\cdot(-1)^3=-1$, it follows that the determinant of $T$ is $-1$ (the determinant can also easily be calculated directly).
It also follows that the characteristic polynomial of $T$ is $(\lambda^2-1)^3(\lambda-1)$; and since $T\ne I$, we must have $T^2=I$ (which can be more readily seen looking at the definition of $T$). So $T^7=T$.