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.