We use the fact that multiplying row $i$ by $\lambda \neq 0$ then adding it to row $j$ does not change the determinant of the matrix.
Also, multiplying row $i$ by $\lambda \neq 0$ also multiples the determinant by $\lambda$. Thus,
\begin{align}
\ & det\pmatrix{1 & 2 & 3 & \dots & n \\ 2 & 3 & 4 & \dots & 1 \\ 3 & 4 & 5 & \dots & 2 \\ \vdots & \vdots & \vdots & & \vdots \\ n & 1 & 2 & \dots & n-1} \\
\ & = \frac{n(n+1)}{2} det\pmatrix{1 & 1 & 1 & \dots & 1 \\ 2 & 3 & 4 & \dots & 1 \\ 3 & 4 & 5 & \dots & 2 \\ \vdots & \vdots & \vdots & & \vdots \\ n & 1 & 2 & \dots & n-1} && \text{as you have} \\
\ & = \frac{n(n+1)}{2} det\pmatrix{1 & 1 & 1 & \dots & 1 & 1 & 1 \\ 0 & 1 & 2 & \dots & n-3 & n-2 & -1 \\ 0 & 1 & 2 & \dots & n-3 & -2 & -1 \\ \vdots & \vdots & \vdots & & \vdots & \vdots & \vdots \\ 0 & 1 & 2-n & \dots & -3 & -2 & -1\\0 & 1-n & 2-n & \dots & -3 & -2 & -1} \\
\ & = \frac{n(n+1)}{2} det\pmatrix{1 & 1 & 1 & \dots & 1 & 1 & 1 \\ 0 & 1 & 2 & \dots & n-3 & n-2 & -1 \\ 0 & 0 & 0 & \dots & 0 & -n & 0 \\ 0 & 0 & 0 & \dots & -n & -n & 0 \\ \vdots & \vdots & \vdots & & \vdots & \vdots & \vdots \\ 0 & 0 & -n & \dots & -n & -n & 0\\0 & -n & -n & \dots & -n & -n & 0} \\
\ & = \frac{n(n+1)}{2} det\pmatrix{1 & 1 & 1 & \dots & 1 & 1 & 1 \\ 0 & 1 & 2 & \dots & n-3 & n-2 & -1 \\ 0 & 0 & 0 & \dots & 0 & -n & 0 \\ 0 & 0 & 0 & \dots & -n & 0 & 0 \\ \vdots & \vdots & \vdots & & \vdots & \vdots & \vdots \\ 0 & 0 & -n & \dots & 0 & 0 & 0\\0 & -n & 0 & \dots & 0 & 0 & 0} \\
\ & = \frac{(-1)^nn^{n-1}(n+1)}{2} det\pmatrix{1 & 1 & 1 & \dots & 1 & 1 & 1 \\ 0 & 1 & 2 & \dots & n-3 & n-2 & -1 \\ 0 & 0 & 0 & \dots & 0 & 1 & 0 \\ 0 & 0 & 0 & \dots & 1 & 0 & 0 \\ \vdots & \vdots & \vdots & & \vdots & \vdots & \vdots \\ 0 & 0 & 1 & \dots & 0 & 0 & 0\\0 & 1 & 0 & \dots & 0 & 0 & 0} \\
\ & = \frac{(-1)^nn^{n-1}(n+1)}{2}\cdot (1 \cdot 1 \cdot 1 \cdot \; \cdots \cdot 1 \cdot -1)(-1)^{\lceil n/2 \rceil +1} \\
\ & = \frac{(-1)^{n+\lceil n/2 \rceil}n^{n-1}(n+1)}{2}
\end{align}
where the second to last line uses the permutation formula for the determinant.
The permutation formula gives the stated result because, as I have explained in the comment section, you have to choose one entry from each row and column and compute their product.
But in the last row, we have to choose the $1$ in order to not get a product of $0$. Similarly with the second to last row, up to the third row.
So far, we have already chosen one entry from each of columns $2$ to $n-1$, thus we need to choose an entry from columns $1$ and $n$ in the remaining two rows. In the second row, we can only choose the $-1$ at the far right to avoid a $0$ product, meaning we can only choose the $1$ in the first column for the first row.
This product is $1\cdot 1\cdot 1\dots 1\cdot -1 \cdot 1=-1$
The parity of this permutation is even if $n=1,2,5,6,9,10,\dots$ and is odd if $n=3,4,7,8,11,12,\dots$, so $sgn (\sigma) = (-1)^{\lceil n/2 \rceil +1}$, giving the determinant of the reduced matrix to be $-1 \cdot (-1)^{\lceil n/2 \rceil +1} = (-1)^{\lceil n/2 \rceil}$