Given an invertible $n\times n$ matrix, an inverse can be obtain through Gaussian elimination. Take an $n\times 2n$ matrix of the form $(A|I_n)$ ($I_n$ is the $n\times n$ identity matrix); row reduce until you get the identity in the first $n$ columns; the last $n$ columns are the inverse. Here: $\begin{align*} \left(\begin{array}{rrrr|rrrr} 1&1&1&1& 1& 0 & 0& 0\\ 1& -1 & 1 & 0 & 0 & 1 & 0 & 0\\ 1 & 1 & 0 & 0& 0 & 0 & 1 & 0\\ 1 & 0 & 0 & 0 & 0 &0 & 0 & 1 \end{array}\right) &\to \left(\begin{array}{rrrr|rrrr} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ 1 & 1 & 0 & 0 & 0 & 0 & 1 & 0\\ 1 & -1 & 1 & 0 & 0 & 1 & 0 & 0\\ 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 \end{array}\right)\\ &\to \left(\begin{array}{rrrr|rrrr} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 & -1\\ 0 & -1 & 1 & 0 & 0 & 1 & 0 & -1\\ 0 & 1 & 1 & 1 & 1 & 0 & 0 & -1 \end{array}\right)\\ &\to \left(\begin{array}{rrrr|rrrr} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 & -1\\ 0 & 0 & 1 & 0 & 0 & 1 & 1 & -2\\ 0 & 0 & 1 & 1 & 1 & 0 & -1 & 0 \end{array}\right)\\ &\to \left(\begin{array}{rrrr|rrrr} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 & -1\\ 0 & 0 & 1 & 0 & 0 & 1 & 1 & -2\\ 0 & 0 & 0& 1 & 1 & -1 & -2 & 2 \end{array} \right) \end{align*}$
So the inverse of your matrix is $\left(\begin{array}{rrrr} 0 & 0 & 0 & 1\\ 0 & 0 & 1 & -1\\ 0 & 1 & 1 & -2\\ 1 & -1 & -2 & 2 \end{array}\right).$
Added. The method can also be used if you don't know whether your matrix is invertible. Perform row reduction; if you get a row that has zeros on the first $n$ columns, then your original matrix was not invertible. Otherwise, you get the identity matrix on the left and obtain an inverse.
Added 2. It is also possible to figure out the inverse in an ad hoc manner because $A$ is fairly simple: interpret it as a linear transformation. The linear transformation maps $e_4$ to $e_1$, so the inverse must map $e_1$ to $e_4$, giving the first column.
$A$ maps $e_3$ to $e_1+e_2$, so the inverse must map $e_1+e_2$ to $e_3$; since it maps $e_1$ to $e_4$, the inverse maps $e_2 = (e_1+e_2)-e_1$ to $e_3-e_4$. That's the second column.
$A$ maps $e_2$ to $e_1-e_2+e_3$, so $A^{-1}$ maps $e_1-e_2+e_3$ to $e_2$; therefore, it maps $e_3 = (e_1-e_2+e_3) +(e_1+e_2) - 2e_1$ to $e_2 + e_3 -2e_4$. This gives the third column.
And finally, since $A$ maps $e_1$ to $e_1+e_2+e_3+e_4$, $A^{-1}$ maps $e_1+e_2+e_3+e_4$ to $e_1$. Hence it maps $e_4= (e_1+e_2+e_3+e_4) - (e_1-e_2+e_3)-2(e_1+e_2)+2e_1$ to $e_1-e_2-2e_3+2e_4$.
Of course, this amounts to just doing the row-reduction in an ad hoc manner...