So the matrix is $A \in M_{n\times n}(\mathbb R)$ and defined:
In row $i$ $1 \le i \le n$ there is $1$ in columns $i+1$ and $i-1$, the rest are $0$'s.
Except for:
$a_{(1,2)}$=1
$a_{(1,n)}$=1
$a_{(n,n-1)}$=1
$a_{(n,1)}$=1
How do I calculate $\det(A)$ using permutations?
Thanks in advance, I'm feeling so lost.
I assume you mean using permutation matrices, as opposed to permutations of the columns.
Consider the matrix $P$ defined by $p_{i+1,i} = 1$ for $i = 1,\dots,n-1$, $p_{1,n} = 1$, and all other entries of $P$ are zero. Note that $P$ is a permutation matrix with $P^n = I$. You should have no trouble finding its complex eigenvalues/eigenvectors (using the minimal polynomial). Note that $P^{-1} = P^T$ (the transpose of $P$).
Now, note that $A = P + P^{-1}$. It follows that $A$ has eigenvalues $\lambda + \lambda^{-1}$ for any eigenvalue $\lambda$ of $P$.
Deduce that the eigenvalues of $A$ are $2 \cos(2 \pi j/n)$ for $j = 0,\dots,n-1$. Thus, the determinant of $A$ is their product $$ \det(A) = 2^n \prod_{j=0}^{n-1} \cos(2 \pi j/n) $$ I am not aware of any further simplification that can be made. Note, however, that $\det(A) = 0$ if and only if $n$ is even.