Characteristic polynomial of checker matrix

Ok! Your conjecture is true.

Let $W$ be the space spanned by the eigenvectors for $\lambda \in \{-3, 3\}$ as described in my comments. Let $V$ be the subspace of $\mathbb{R}^{4n}$ consisting of vectors of the form

$$V = \{(a,b,a,b,a,b, \ldots, a, x, y, b, a, b, \ldots, a,b)\},$$

where the entries corresponding to $x,y$ are in positions $2n$ and $2n+1$ of the vector. (So $V$ is the orthogonal complement of $W$.)

Let $T : V \to \mathbb{R}^4$ by $T(\vec{v}) = (a,b,x,y)$ in the obvious way (so $T$ is an isomorphism).

We can check that $V$ is invariant under the action of $C_{n}$. And moreover, that $$T \circ C_{n} \circ T^{-1} \begin{pmatrix}a\\b\\x\\y \end{pmatrix} = \begin{pmatrix}(2n-1)a +y + 3b\\(2n-1)b+x+3a\\(2n-1)b+x\\(2n-1)a+y \end{pmatrix}.$$

Thus, $C_{n}$ restricted to $V$ is isomorphic to the above linear map on $\mathbb{R}^4$, namely

$$\begin{pmatrix}a\\b\\x\\y\end{pmatrix} \mapsto \begin{pmatrix}2n-1 & 3 & 0 &1\\ 3 & 2n-1 & 1 &0\\ 0 & 2n-1 & 1 &0\\ 2n-1 & 0 & 0 &1\end{pmatrix} \begin{pmatrix}a\\b\\x\\y\end{pmatrix},$$

and this map has the desired remaining four eigenvalues as in your conjecture.