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.