Let $A$ be the $N\times N$ skew-symmetric matrix defined by the rules $$\begin{cases}A_{ij}=1 & \text{when }i<j, \\ A_{ii}=0 & \forall i=1,...,N \\ A_{ij}=-1 &\text{when }i>j.\end{cases}$$ Can we derive a rule to write the eigenvalues and eigenvectors of this matrix as a function of $N$? What about a rule for determining its exponential?
For low values of $N$, it seems manageable: for example if $N=2$ we have the conjugate pair of eigenvalues $\lambda=i, \lambda^*=-i$, while for $N=3$ the eigenvalues are $\{0, \pm i\sqrt3\}$. Numerically, things seem to get complicated as soon as $N\ge 7$ however.
I base my answer on the comment of user1551 so thanks to them for the idea.
Transforming $A$ to a circulant matrix
Define $\alpha= \sqrt[N]{-1}$ $(= \exp(\pi i /N) )$ and $D = \mathrm{diag} (\alpha^1,\alpha^2 \dots ,\alpha^N )$. Note the slight difference to the definition in the comments. Then $C = D^{-1} A D$ is a circulant matrix.
For example for $N=3$ we have $$ C = \left[\begin{matrix}0 & \sqrt[3]{-1} & \left(-1\right)^{\frac{2}{3}}\\\left(-1\right)^{\frac{2}{3}} & 0 & \sqrt[3]{-1}\\\sqrt[3]{-1} & \left(-1\right)^{\frac{2}{3}} & 0\end{matrix}\right]. $$ And in general the first column $c$ of $C$ (which defines $C$ uniquely) is given by $$ c(j) = \begin{cases} 0, \quad \text{if $j=1$,}\\ \exp( \pi i (N-j+1) /N), \quad \text{else}. \end{cases} $$ Define the Fourier matrix $\mathcal{F} \in \mathbb{C}^{N \times N}$ by $\mathcal{F}(j,k)= \exp(-2\pi i jk /N ) $.
Then we have $$C= \mathcal{F}^{-1} \mathrm{diag}(\mathcal{F}c) \mathcal{F}, \tag{1}$$ see here (note the wrong factor of $1/N$ on wikipedia).
Eigenvalues
From equation $(1)$ we obtain the eigenvalues $\lambda_j$ as \begin{align} \lambda_j& = \sum_{k=1}^N \mathcal{F}(j,k) c(k) = \sum_{k=2}^N \exp(-2 \pi i (j-1)(k-1) /N ) \exp( \pi i (N-k+1) /N) \\ &= \sum_{k=1}^{N-1} \exp( \pi i [k(1-2j)+N]/N). \end{align}
In the case $N=3$ we obtain: \begin{align*} \lambda_1 &= \exp(2\pi i/3 )+ \exp(\pi i/3) = i \sqrt{3}, \\ \lambda_2 &= 1+ \exp(-\pi i) = 0, \\ \lambda_3 &= \exp(-2\pi i/3 )+ \exp(-7\pi i/3) = - i \sqrt{3}, \\ \end{align*} which agrees with the result obtained using the usual methods.
I also numerically compared the formula with results for larger $N$ and it seems to give the correct results.
Eigenvectors
From the above we have
$$A = D C D^{-1} = D \mathcal{F}^{-1} \mathrm{diag}(\mathcal{F}c) \mathcal{F} D^{-1}.$$ Therefore the eigenvector $v_j$ corresponding to $\lambda_j$ is given as the $j$-th column of $D \mathcal{F}^{-1} $. We have $\mathcal{F}^{-1} (j,k) =1/N \mathcal{F}^*(j,k) = 1/N\exp(2\pi i jk /N )$ and therefore $$ \begin{align*} v_j (k) &= [D \mathcal{F}^{-1} ](k,j) = \sum_{l=1}^N D(k,l) \mathcal{F}^{-1}(l,j) \\ &=1/N \exp(\pi i k/N) \exp(2\pi i (k-1)(j-1) /N )\\ &= 1/N\exp(\pi i (2kj-2j+2-k) /N ) \\ &= 1/N\exp(\pi i (2j(k-1)+2-k) /N ). \end{align*} $$ For $N=3$ we obtain $$ \begin{align*} v_1 = 1/3 \begin{pmatrix} \exp(\pi i 1 /3 ) \\ \exp(\pi i 2 /3 ) \\ \exp(\pi i 3/3 ) \\ \end{pmatrix}, \ v_2 = 1/3 \begin{pmatrix} \exp(\pi i 1 /3 ) \\ \exp(\pi i 4/3 ) \\ \exp(\pi i 7/3 ) \\ \end{pmatrix}, \ v_3 = 1/3 \begin{pmatrix} \exp(\pi i 1 /3 ) \\ \exp(\pi i 6 /3 ) \\ \exp(\pi i 11/3 ) \\ \end{pmatrix}, \end{align*} $$ which can be checked to be the eigenvectors of $A$ to the eigenvalues found above.
I also verified this formula numerically for higher $N$ and it seems to work.
The matrix exponential can also be obtained in closed form from this. I leave the details to you.