Determinant of a special skew-symmetric matrix

Here is a combinatorial way to answer this. If we have a skew-symmetric matrix $A=\{a_{ij}\}_{1\le i,j\le 2n}$, then $\det(A)=Pf(A)^2$, where $Pf(A)$ is the Pfaffian of $A$. We know from standard methods that $$Pf(A)=\sum_{\pi \in \Pi}\text{sgn}(\pi)a_{\pi(1),\pi(2)}\cdots a_{\pi(2n-1),\pi(2n)}$$ where $\Pi$ is the set of permutations $\pi\in S_{2n}$ which satisfy $\pi(2k-1)<\pi(2k)$ for $1\le k\le n$ and $\pi(1)\le \pi(3)\le \cdots \le \pi(2n-1)$. In our case all $a_{ij}$ with $i < j$ have the same value $-1$, so we only need to prove that $$|\sum_{\pi \in \Pi}\text{sgn}(\pi)|=1.$$ To do this we will exhibit an involution on $\Pi\backslash\{id\}$ (the permutations in $\Pi$ that are not the identity).

Let $\pi \in \Pi\backslash\{id\}$, there will be a smallest $k$ so that $\pi(2k-1)= \pi(2k+1)-1$. define $\pi'$ to be the same as $\pi$ but with $\pi'(2k)=\pi(2k+2)$ and $\pi'(2k+2)=\pi(2k)$. I will leave it as an exercise for you to prove that $\pi'\in \Pi\backslash\{id\}$, $\pi''=\pi$ and that $\text{sgn}(\pi')=-\text{sgn}(\pi)$ so that $$\sum_{\pi \in \Pi}\text{sgn}(\pi)=\text{sgn}(id)=1.$$

Let $$P=\begin{pmatrix}1\\-1&1\\&-1&1\\&&\ddots&\ddots\\&&&-1&1\end{pmatrix}.$$ Then $$PA_{2n}=\begin{pmatrix}0&1&1&\ldots&1\\-1&-1\\&-1&-1\\&&\ddots&\ddots\\&&&-1&-1\end{pmatrix}.$$ Computing by row expansion, we get $\det(PA_{2n}) = 0 - (-1) + (-1) - \ldots + (-1) - (-1) = 1$. Since $\det(P)=1$, we are now done.

Edit: By considering $PA_nP^{-1}$, actually we can further show that the characteristic polynomial of $A_n$ is $p(\lambda)=\det(\lambda I_n-A_n)=\frac12\left((\lambda+1)^n+(\lambda-1)^n\right)$, regardless of whether $n$ is even or odd. Therefore, if $n$ is even and $\lambda$ is a (necessarily nonzero) eigenvalue of $A_n$, so is $1/\lambda$.

Based upon J.M.'s comments, I'd like to approach this in a different way from Davide's answer. Starting from a slightly different partitioning $$A_{2n+2} = \left(\begin{array}{cc} A_2 & B \\ -B^T & A_{2n}\end{array}\right)$$ where $B$ is a $2\times 2n$ matrix with all entries set to $1$, we know

$$\det(A_{2n+2}) = \det(A_2)\det(A_{2n} + B^T A_2^{-1} B).$$

Inspection reveals that $A_2^{-1} = A_2^T$, so

$$A_2^{-1} B = \left( \begin{array}{rrc} -1 & -1 & \ldots \\ 1 & 1 & \ldots \end{array} \right)$$

which means $B^T A_2^{-1} B = 0$. Hence, $\det(A_{2n+2}) = \det(A_{2n})\det(A_2)$. Since we know the determinant of $A_{2n}$ for $n=1\ \text{and}\ 2$ is $1$, clearly $\det(A_{2n}) = 1 \ \forall\ n \ge 1$.

Edit: it occurs to me that the inductive step is simplified by recognizing that $\det(A_{2n+2}) = \det(A_{2n})$ because $\det(A_2) = 1$. Then, this implies $\det(A_{2n}) = \det(A_2)$.