A simple brute force method worked (even though I'm not happy with this).

Let $\zeta=\xi+i\eta$ be a non-positive eigenvalue of $M$ and $\left[\begin{matrix} x & y & z \end{matrix}\right]^T$ be a corresponding eigenvector. This gives equations \begin{align} Ax + By \qquad &= \zeta x \\ B^Tx + Cy - B^Tz &= \zeta y\\ -Ax -By + Az &= \zeta z \end{align} From the first and the third, one obtains $$z = -\zeta(\zeta-A)^{-1}x \mbox{ and } x-z = (1+\zeta(\zeta-A)^{-1})x.$$ To ease the notation, put $A(\zeta):=1+\zeta(\zeta-A)^{-1}=(2\zeta-A)(\zeta-A)^{-1}$. From the second, $$y = (\zeta-C)^{-1}B^T(x-z) = (\zeta-C)^{-1}B^T A(\zeta) x.$$ By combining with the first, one obtains $$(\zeta-A)x = By = B(\zeta-C)^{-1}B^T A(\zeta)x.$$ Note that the imaginary part of $(\zeta- C)^{-1}$ is $$\Im\frac{1}{\zeta- C}=\Im\frac{\bar{\zeta}-C}{|\zeta-C|^2} =-\eta\frac{1}{|\zeta-C|^2}.$$ Take the inner product with $A(\zeta)x$ and look at the imaginary part: $$\Im \langle (\zeta-A)x,A(\zeta)x\rangle = -\eta \langle B|\zeta-C|^{-2}B^TA(\zeta)x,A(\zeta)x\rangle.$$ Now, since \begin{align*} A(\bar{\zeta})(\zeta-A) &= \frac{(2\bar{\zeta}-A)(\zeta-A)^2}{|\zeta-A|^2} \\ &=\frac{2|\zeta|^2\zeta-4|\zeta|^2A+2\bar{\zeta}A^2-\zeta^2A+2\zeta A^2-A^3}{|\zeta-A|^2}, \end{align*} one obtains $$2\eta\langle \frac{|\zeta|^2-\xi A }{|\zeta-A|^2}x,x\rangle =-\eta \langle B|\zeta-C|^{-2}B^TA(\zeta)x,A(\zeta)x\rangle.$$ Hence, unless $\eta=0$, $$\xi\geq\frac{|\zeta|^2}{\|A\|}>0.$$ Let's deal with the case $\eta=0$. Suppose for a contradiction that $\xi\le0$. Then \begin{align*} \langle A(\xi)(-\xi+A)x,x\rangle &= \langle B(-\xi+C)^{-1}B^T A(\xi)x,A(\xi)x\rangle\\ &\le \langle BC^{-1}B^T A(\xi)x,A(\xi)x\rangle\\ &< \langle A A(\xi)x,A(\xi)x\rangle, \end{align*} but this is in contradiction with the fact that $A(\xi)\succ0$ and $-\xi + A\succeq A(\xi)A$.

In principle, one can find a lower bound for $\xi$, but I'm too tired for this.

A simple brute force method worked (even though I'm not happy with this).

Let $\zeta=\xi+i\eta$ be a non-positive eigenvalue of $M$ and $\left[\begin{matrix} x & y & z \end{matrix}\right]^T$ be a corresponding eigenvector. This gives equations \begin{align} Ax + By \qquad &= \zeta x \\ B^Tx + Cy - B^Tz &= \zeta y\\ -Ax -By + Az &= \zeta z \end{align} From the first and the third, one obtains $$z = -\zeta(\zeta-A)^{-1}x \mbox{ and } x-z = (1+\zeta(\zeta-A)^{-1})x.$$ To ease the notation, put $A(\zeta):=1+\zeta(\zeta-A)^{-1}=(2\zeta-A)(\zeta-A)^{-1}$. From the second, $$y = (\zeta-C)^{-1}B^T(x-z) = (\zeta-C)^{-1}B^T A(\zeta) x.$$ By combining with the first, one obtains $$(\zeta-A)x = By = B(\zeta-C)^{-1}B^T A(\zeta)x.$$ Note that the imaginary part of $(\zeta- C)^{-1}$ is $$\Im\frac{1}{\zeta- C}=\Im\frac{\bar{\zeta}-C}{|\zeta-C|^2} =-\eta\frac{1}{|\zeta-C|^2}.$$ Take the inner product with $A(\zeta)x$ and look at the imaginary part: $$\Im \langle (\zeta-A)x,A(\zeta)x\rangle = -\eta \langle B|\zeta-C|^{-2}B^TA(\zeta)x,A(\zeta)x\rangle.$$ Now, since \begin{align*} A(\bar{\zeta})(\zeta-A) &= \frac{(2\bar{\zeta}-A)(\zeta-A)^2}{|\zeta-A|^2} \\ &=\frac{2|\zeta|^2\zeta-4|\zeta|^2A+2\bar{\zeta}A^2-\zeta^2A+2\zeta A^2-A^3}{|\zeta-A|^2}, \end{align*} one obtains $$2\eta\langle \frac{|\zeta|^2-\xi A }{|\zeta-A|^2}x,x\rangle =-\eta \langle B|\zeta-C|^{-2}B^TA(\zeta)x,A(\zeta)x\rangle.$$ Hence, unless $\eta=0$, $$\xi\geq\frac{|\zeta|^2}{\|A\|}>0.$$ Let's deal with the case $\eta=0$. Suppose for a contradiction that $\xi\le0$. Then \begin{align*} \langle A(\xi)(-\xi+A)x,x\rangle &= \langle B(-\xi+C)^{-1}B^T A(\xi)x,A(\xi)x\rangle\\ &\le \langle BC^{-1}B^T A(\xi)x,A(\xi)x\rangle\\ &< \langle A A(\xi)x,A(\xi)x\rangle, \end{align*} but this is in contradiction with the fact that $A(\xi)\succ0$ and $-\xi + A\succeq A(\xi)A$.

ADDED: We assume $\left[\begin{smallmatrix} A_0 & B_0 & \\ B_0^T & C_0\end{smallmatrix}\right]\succ 2\epsilon$ and will show that any eigenvalue of $\left[\begin{smallmatrix} A_0 & B_0 & \\ B_0^T & C_0 & -B_0^T \\ -A_0 & -B_0 & A_0 \end{smallmatrix}\right]$ has its real part at least $\epsilon$. To ease notation, we consider $A=A_0-2\epsilon$ and $C=C_0-2\epsilon$ instead of $A_0$ and $C_0$. By the IVT trick, it suffices to show there is no solution for $$\left[\begin{matrix} A+2\epsilon & B & \\ B^T & C+2\epsilon & -B^T \\ -(A+2\epsilon) & -B & A+2\epsilon \end{matrix}\right] \left[\begin{matrix} x \\ y \\ z \end{matrix}\right] = (\epsilon+i\eta) \left[\begin{matrix} x \\ y \\ z \end{matrix}\right],$$ with $\left[\begin{smallmatrix} A & B & \\ B^T & C\end{smallmatrix}\right]\succ 0$, $\epsilon>0$, $\eta\in\mathbb{R}$, and $\left[\begin{smallmatrix} x & y & z \end{smallmatrix}\right]\neq 0$.

The first row + the third row: $-\epsilon x + (A+\epsilon) z = i\eta (x+z).$ Hence $$x-z=(1-\frac{\epsilon+i\eta}{A+\epsilon-i\eta})x =\frac{A-2i\eta}{A+\epsilon-i\eta}x=:A(\eta)x.$$ Together with the second row: $$y=-(C+\epsilon-i\eta)^{-1}B^T(x-z)=-(C+\epsilon-i\eta)^{-1}B^TA(\eta)x.$$ It follows that $x\neq0$. Together with the first row: \begin{align*} (A+\epsilon -i\eta) x = -By = B(C+\epsilon-i\eta)^{-1}B^TA(\eta)x \end{align*} and so \begin{equation}\tag{$\ast$} \langle (A+\epsilon -i\eta) x,A(\eta)x\rangle = \langle B(C+\epsilon-i\eta)^{-1}B^TA(\eta)x, A(\eta)x\rangle.\end{equation} Do some calculations: $$A(\eta)^*(A+\epsilon -i\eta) = |A+\epsilon-i\eta|^{-2}(A+\epsilon-i\eta)^2(A+2i\eta),$$ $$(A+\epsilon-i\eta)^2(A+2i\eta) = A^3 + 2 \epsilon A^2 + \epsilon^2 A + 3 \eta^2 A + 4\epsilon \eta^2 + 2i\eta (\epsilon A + \epsilon^2 - \eta^2),$$ $$A(\eta)^*B(C+\epsilon-i\eta)^{-1}B^TA(\eta) =A(\eta)^*B\frac{C+\epsilon +i\eta}{|C+\epsilon-i\eta|^2}B^TA(\eta),$$ $$A(\eta)^*B\frac{C+\epsilon}{|C+\epsilon-i\eta|^2}B^TA(\eta) \preceq A(\eta)^*B C^{-1} B^T A(\eta) \prec A|A(\eta)|^2,$$ $$A|A(\eta)|^2 = |A+\epsilon-i\eta|^{-2}(A^3+4\eta^2A).$$ Look at the real part of $(\ast)$: for $w=|A+\epsilon-i\eta|^{-1}x$, $$\langle(2\epsilon A^2+\epsilon^2A+4\epsilon\eta^2)w,w\rangle < \langle \eta^2Aw,w\rangle.$$ Look at the imaginary part of $(\ast)$: as $\eta\neq0$ from the previous inequality, $$\langle(\epsilon A + \epsilon^2 - \eta^2)w,w\rangle =\langle A(\eta)^*B\frac{1}{|C+\epsilon-i\eta|^2}B^TA(\eta)x,x\rangle/2 \geq0.$$ Combine the last two: \begin{align*} 2\epsilon \|Aw\|^2+\epsilon^2\langle Aw,w\rangle +4\epsilon\eta^2\|w\|^2 &< \eta^2\langle Aw,w\rangle\\ &\le (\epsilon \frac{\langle Aw,w\rangle}{\|w\|^2} +\epsilon^2)\langle Aw,w\rangle\\ &\le\epsilon\|Aw\|^2 +\epsilon^2\langle Aw,w\rangle. \end{align*} We arrive at a contradiction.