Fourier transform does it.

Denote by $u_j$ $(j=0,1,\ldots,n-1$) the column-vector with coordinates $(x_{ji})_{1\leqslant i\leqslant n-1}$. Note that $u_0+u_1+\ldots+u_{n-1}=0$ and any $n-1$ vectors $u_i$'s are linearly independent. The idea is to write the matrix $C_1=AB_1$ in the basis $\{u_0,u_1,\ldots,u_{n-1}\}\setminus \{u_{(n-1)/2}\}$. It has a form $\pmatrix{X&Y\\0&Z}$, where $X,Z$ are lower-triangular, thus its eigenvectors are diagonal elements of $X$ and $Z$, and these are exactly $-\frac{n-1}{2},\cdots,-1,1,\cdots,\frac{n-1}{2}$. See details below.

Lemma 1. For $\ell\in \{1,2,\ldots,n\}$ we have $$\sum_{j=1}^{n-1}\frac{x_{j\ell}}{x_j-1}=\frac{n-1}2-\ell+1.$$

Proof. For $1\leqslant j\leqslant n-1$ we have $$(x_j-1)((n-1)+(n-2)x_j+(n-3)x_{2j}+\ldots+x_{(n-2)j})\\=(1+x_j+x_{2j}+\ldots+x_{(n-1)j})-n=-n.$$ Therefore $$ \sum_{j=1}^{n-1}\frac{x_{j\ell}}{x_j-1}=-\frac1n\sum_{j=1}^{n-1}x_{j\ell} \sum_{s=1}^{n}(s-1)x_{j(n-s)}=\frac{n-1}2-\frac1n\sum_{j=0}^{n-1}x_{j\ell} \sum_{s=1}^{n}(s-1)x_{j(n-s)}\\ =\frac{n-1}2-\frac1n\sum_{s=1}^{n} \sum_{j=0}^{n-1}(s-1)x_{j(n-s+\ell)}=\frac{n-1}2-\ell+1, $$ since $\sum_{j=0}^{n-1}x_{j(n-s+\ell)}=n\cdot \delta_{s-\ell}$. $\square$

Lemma 2. For $p=0,1,\ldots,n-2$ we have $$C_1u_p=\left(\frac{n-1}2-p\right)u_p-\left(\frac{n-1}2-p-1\right)u_{p+1}.$$ Also, for $p=n-1$ we get $$C_1u_{n-1}=-\frac{n-1}2u_{n-1}-\frac{n-1}2 u_0.$$

Proof. For $p=0,1,\ldots,n-2$ we have $$ \left[C_1u_p\right]_i=\sum_{1\leqslant j\leqslant n-1,j\ne i} \frac{1-x_j}{1-x_{i-j}}x^{pj}=\sum_{1\leqslant j\leqslant n-1,j\ne i} \frac{x_{(p+1)j}-x_{(p+2)j}}{x_j-x_{i}}\\ =\sum_{0\leqslant j\leqslant n-1,j\ne i} \frac{x_{(p+1)j}-x_{(p+2)j}}{x_j-x_{i}}=x_{pi} \sum_{0\leqslant j\leqslant n-1,j\ne i} \frac{x_{(p+1)(j-i)}}{x_{j-i}-1} -x_{(p+1)i}\sum_{0\leqslant j\leqslant n-1,j\ne i} \frac{x_{(p+2)(j-i)}}{x_{j-i}-1}\\ =\left(\frac{n-1}2-p\right)x_{pi}-\left(\frac{n-1}2-p-1\right)x_{(p+1)i} $$ by Lemma 1. For $p=n-1$ the last coefficient of $x_{(p+1)i}$ corresponds to the case $\ell=1$ in Lemma 1 and therefore equals $-(\frac{n-1}2-1+1)=-\frac{n-1}2$. $\square$

So, we proved the aforementioned block representation of $C_1$ in the basis $\{u_0,u_1,\ldots,u_{n-1}\}\setminus \{u_{(n-1)/2}\}$.

Fourier transform does it.

Denote by $u_j$ $(j=0,1,\ldots,n-1$) the column-vector with coordinates $(x_{ji})_{1\leqslant i\leqslant n-1}$. Note that $u_0+u_1+\ldots+u_{n-1}=0$ and any $n-1$ vectors $u_i$'s are linearly independent. The idea is to write the matrix $C_1=AB_1$ in the basis $\{u_0,u_1,\ldots,u_{n-1}\}\setminus \{u_{(n-1)/2}\}$. It has a form $\pmatrix{X&Y\\0&Z}$, where $X,Z$ are lower-triangular, thus its eigenvalues are diagonal elements of $X$ and $Z$, and these are exactly $-\frac{n-1}{2},\cdots,-1,1,\cdots,\frac{n-1}{2}$. See details below.

Lemma 1. For $\ell\in \{1,2,\ldots,n\}$ we have $$\sum_{j=1}^{n-1}\frac{x_{j\ell}}{x_j-1}=\frac{n-1}2-\ell+1.$$

Proof. For $1\leqslant j\leqslant n-1$ we have $$(x_j-1)((n-1)+(n-2)x_j+(n-3)x_{2j}+\ldots+x_{(n-2)j})\\=(1+x_j+x_{2j}+\ldots+x_{(n-1)j})-n=-n.$$ Therefore $$ \sum_{j=1}^{n-1}\frac{x_{j\ell}}{x_j-1}=-\frac1n\sum_{j=1}^{n-1}x_{j\ell} \sum_{s=1}^{n}(s-1)x_{j(n-s)}=\frac{n-1}2-\frac1n\sum_{j=0}^{n-1}x_{j\ell} \sum_{s=1}^{n}(s-1)x_{j(n-s)}\\ =\frac{n-1}2-\frac1n\sum_{s=1}^{n} \sum_{j=0}^{n-1}(s-1)x_{j(n-s+\ell)}=\frac{n-1}2-\ell+1, $$ since $\sum_{j=0}^{n-1}x_{j(n-s+\ell)}=n\cdot \delta_{s-\ell}$. $\square$

Lemma 2. For $p=0,1,\ldots,n-2$ we have $$C_1u_p=\left(\frac{n-1}2-p\right)u_p-\left(\frac{n-1}2-p-1\right)u_{p+1}.$$ Also, for $p=n-1$ we get $$C_1u_{n-1}=-\frac{n-1}2u_{n-1}-\frac{n-1}2 u_0.$$

Proof. For $p=0,1,\ldots,n-2$ we have $$ \left[C_1u_p\right]_i=\sum_{1\leqslant j\leqslant n-1,j\ne i} \frac{1-x_j}{1-x_{i-j}}x^{pj}=\sum_{1\leqslant j\leqslant n-1,j\ne i} \frac{x_{(p+1)j}-x_{(p+2)j}}{x_j-x_{i}}\\ =\sum_{0\leqslant j\leqslant n-1,j\ne i} \frac{x_{(p+1)j}-x_{(p+2)j}}{x_j-x_{i}}=x_{pi} \sum_{0\leqslant j\leqslant n-1,j\ne i} \frac{x_{(p+1)(j-i)}}{x_{j-i}-1} -x_{(p+1)i}\sum_{0\leqslant j\leqslant n-1,j\ne i} \frac{x_{(p+2)(j-i)}}{x_{j-i}-1}\\ =\left(\frac{n-1}2-p\right)x_{pi}-\left(\frac{n-1}2-p-1\right)x_{(p+1)i} $$ by Lemma 1. For $p=n-1$ the last coefficient of $x_{(p+1)i}$ corresponds to the case $\ell=1$ in Lemma 1 and therefore equals $-(\frac{n-1}2-1+1)=-\frac{n-1}2$. $\square$

So, we proved the aforementioned block representation of $C_1$ in the basis $\{u_0,u_1,\ldots,u_{n-1}\}\setminus \{u_{(n-1)/2}\}$.