Call that matrix $A$ and notice that it is a permutation of a circulant matrix,
$$
A = CP
$$
Where $P$ is a permutation matrix with ones on the anti-diagonal, and zeros in all other positions. Then
$$
\det[A] = \det[CP] = \det[C]\det[P]
$$
The determinant of the permutation part can be shown to depend on the size $n$. It can be written as
$$
\det[P] = (-1)^{\left\lfloor\frac{n}{2}\right\rfloor}
$$
Now $C$ is
$$
\begin{bmatrix}
a_{n} & a_{n-1} & a_{n-2} & \cdots & a_1\\
a_{1} & a_{n} & a_{n-1} & \cdots & a_2\\
a_{2} & a_{1} & a_n & \cdots & a_3\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
a_{n-1} & a_{n-2} & a_{n-3} & \cdots & a_{n}\\
\end{bmatrix}
$$
$C$ is a circulant matrix. Define the associated polynomial
$$
f(\omega) = a_n + \sum_{k=1}^{n-1} a_k\omega^k
$$
Then using the product formula on the Wikipedia page for circulant matrices,
$$
\det[C] = \prod_{j=0}^{n-1}f(\omega_j),
$$
where $\omega_j=e^{\frac{2\pi i j}{n}}$ and $i=\sqrt{-1}$. Then the final formula is
$$
\det[A] = (-1)^{\left\lfloor\frac{n}{2}\right\rfloor}\prod_{j=0}^{n-1}\left(a_n + \sum_{k=1}^{n-1} a_k\omega_j^k\right)
$$
The eigenvalues and eigenvectors can be found exactly. Let the number of block rows in $M$ be $K$. Let's write the eigenproblem as $MX = \lambda X$ where $X$ is a block vector
$$
X = \begin{pmatrix}
x_1\\
x_2\\
\vdots\\
x_K
\end{pmatrix}
$$
Each block row eigenvalue equation can now be written as
$$Ax_k + A^\top x_{k+1} + x_{k-1} = \lambda x_k, \quad k= 1, \dots, K.$$
Here $x_k$ are assumed to be periodical, so $x_{K+1} \equiv x_1$ and $x_0 \equiv x_K$. Each $x_k$ is a vector of length $n$.
Let's apply discrete Fourier transform, just like it is done for regular circulant matrices.
$$
x_k = \sum_{m=0}^{K-1} \omega^{m(k-1)} z_m
$$
Here $\omega = \exp \frac{2 \pi i} {K}$ and each $z_m$ is a vector of length $n$.
Let's call
$$X^{(m)} = \begin{pmatrix}
z_m\\
\omega^m z_m\\
\vdots\\
\omega^{m(k-1)} z_m\\
\vdots\\
\omega^{m(K-1)} z_m
\end{pmatrix}
= f_m \otimes z_m.
$$
the $m$-th harmonic of the solution $X$. Here $f_m$ is the $m$-th column of the $DFT$-like matrix and $\otimes$ denotes Kronecker product.
It is obvious that
$$
X = X^{(0)} + \dots + X^{(K-1)}.
$$
I state that all eigenvectors $X$ of the original problem can be found as pure harmonics, that is all $X^{(m)} = 0$ except for some $m = m_0$. Harmonics are linearly independent since they are orthogonal:
$$
(X^{(m)}, X^{(m')}) = (f_m \otimes z_m, f_{m'} \otimes z_{m'}) = (f_m, f_{m'}) (z_m, z_{m'}) = K \delta_{mm'} (z_m, z_{m'}).
$$
Each harmonic $X^{(m)}$ gives $n$ eigenvalues govern by
$$
[A + \omega^m A^\top + \omega^{-m}] z_m = \lambda z_m. \tag{*}
$$
We may introduce $B_m = \omega^{-m/2} A + \omega^{m/2} A^\top$ which for real matrices $A$ is hermitian.
$$
B_m z_m = \mu z_m, \quad \mu \in \mathbb R. \tag{**}
$$
Eigenvalues of (*) and (**) are related by
$$
\lambda = \omega^{-m} + \omega^{m/2} \mu.
$$
This is probably best we can do for the second question.
For the $A$ being upper shift matrix
we may proceed.
$$
B_m = \begin{pmatrix}
0 & \omega^{-m/2} \\
\omega^{m/2} & 0 & \omega^{-m/2} \\
&\ddots&\ddots&\ddots\\
&&\omega^{m/2} & 0 & \omega^{-m/2} \\
&&&\omega^{m/2} & 0
\end{pmatrix}
$$
Let's introduce $q = \omega^{m/2} = \exp \frac{\pi i m}{K}$. Then $\omega^{-m/2} = q^{-1} = \bar q$.
Again, rewriting the eigenproblem $B_m u = \mu u$ as a tridiagonal system of equations we get
$$
u_0 = 0\\
q u_{p-1} + \bar q u_{p+1} = \mu u_p, \qquad p = 1, \dots, n\\
u_{n+1} = 0.
$$
Plugging $u_p = e^{i \alpha p}$ as a general solution for the middle equations we get
$$
q e^{-i \alpha} + \bar q e^{i \alpha} = \mu \implies
\mu = 2 \cos (\alpha - \arg q) = 2 \cos \left(\alpha - \frac{m \pi}{K}\right)
$$
Note that sole $e^{i \alpha p}$ cannot satisfy the boundary conditions $u_0 = u_{n+1} = 0$. We might combine two $e^{i \alpha p} - e^{i \alpha' p}$ with different $\alpha$ provided that $\mu_\alpha = \mu_{\alpha'}$. Let's take
$$
\alpha - \frac{m \pi}{K} = -\alpha' + \frac{m \pi}{K} \implies
\alpha' = -\alpha + \frac{2 m \pi}{K}.
$$
Satisfying $u_{n+1} = 0$ gives an equation for $\alpha$
$$
0 = u_{n+1} = e^{i\alpha (n+1)} - e^{i \alpha' (n+1)} = e^{i \alpha' (n+1)} \left(
e^{i (\alpha - \alpha') (n+1)} - 1
\right).
$$
$$
\left(2\alpha - \frac{2m\pi}{K} \right) (n + 1) = 2\pi d, \qquad d = 1, \dots, n\\
\alpha = \frac{m \pi}{K} + \frac{\pi d}{n + 1}.
$$
This gives $n$ solutions for each of $K$ harmonics ($m = 0, \dots, K-1; d = 1,\dots,n$):
$$
\alpha_{m,d} = \frac{m \pi}{K} + \frac{\pi d}{n + 1}\\
\mu_{m,d} = 2\cos \left(\frac{\pi d}{n + 1}\right)\\
\lambda_{m,d} = \omega^{-m} + \omega^{m/2} \mu_{m,d}\\
(z_{m,d})_p = \exp \left(i \alpha_{m,d} p\right) - \omega^{mp}\exp \left(-i \alpha_{m,d} p\right)
$$
Here's a small verification in Python.
I can't hold myself from posting a plot of the eigenvalues for $K=16, n=24$. The eigenvalues are contained in a deltoid.
Best Answer
Let's address the easy part first: if the lower-left entry is a $0$ instead of $B$, then $C$ is block upper-triangular, which means that its determinant is given by the product of the determinants of the diagonal blocks. That is, we would find $$ \det(C) = \det(A)^n = (a^2 - b^2)^n. $$ In light of my below discussion of the original matrix: the eigenvalues of $A$ are of the form $a \pm b$. Correspondingly, $C$ has the same eigenvalues, but each with multiplicity $n$.
The matrix $C$ (as originally presented) can be written nicely in terms of the Kronecker product. In particular, let $J$ and $P$ denote the matrices (of sizes $2 \times 2$ and $n \times n$ respectively) $$ J = \pmatrix{0&1\\1&0}, \quad P = \pmatrix{0&1\\&0&1\\&&\ddots&\ddots\\ &&&0&1\\ 1&&&&0}, $$ and let $I_k$ denote the identity matrix of size $k$. Then $$ C = aI + b(I_n \otimes J) + c(P \otimes I_2). $$ The eigenvalues of $J$ are $\pm 1$, and the eigenvalues of $P$ are the $n$th roots of unity (i.e. $e^{2\pi ki/n}$ for $k = 0,\dots,n-1$).
The properties of the Kronecker product allow us to deduce that for matrices $X,Y$ with eigenvalues $\lambda_1,\dots,\lambda_m$ and $\mu_1,\dots,\mu_n$ respectively, it can be shown that the eigenvalues of $I_n\otimes X + Y\otimes I_m$ are given by $\mu_j + \lambda_k$ for all pairs $j,k$ with $1 \leq j \leq m$ and $1 \leq k \leq n$. Applying this to $X = bJ$ and $y = cP$ allows us to see that the eigenvalues of $C - aI$ are given by $$ (-1)^jb + e^{2\pi k i/n} c, \quad 0 \leq j \leq 1, \quad 0 \leq k \leq n-1. $$ Thus, the eigenvalues of $C$ have the form $$ a + (-1)^jb + e^{2\pi k i/n} c, \quad 0 \leq j \leq 1, \quad 0 \leq k \leq n-1. $$ The determinant of $C$ is the product of its eigenvalues.
If $a,b,c$ are real, it is notable that the eigenvalues of $C$ can be put into complex conjugate pairs. For $k = 0,1,\dots,\lfloor (n-1)/2\rfloor$, we find that $$ (a + (-1)^jb + e^{2\pi k i/n} c)(a + (-1)^jb + e^{2\pi (n-k) i/n} c) = \\ |a + (-1)^jb + e^{2\pi k i/n} c|^2 = \\ [(a + (-1)^jb) + c\cos(2\pi k/n)]^2 + c\sin^2(2\pi k/n). $$ With that, the determinant of $C$ can be written as the product of real numbers.