The system is not at rest. If you consider the masses and the pulley to be one system, you can understand the behaviour of the system by the behaviour of its centre of mass. Unless the masses are equal, the centre of mass of the system is not at rest.
It might be useful to think of it in this way - Inside the system boundary mass $m_1$ moves down through a distance while mass $m_2$ moves up by the same distance. So, the centre of mass has moved down (or up depending whether $m_1 > m_2$).
So, the tension would be given by the equation:
$$(m_1+m_2)a_{cm} = (m_1+m_2)g - T_c $$
You can further work out that
$a_{cm} = a(m_2-m_1) /(m_1+m_2)$, where a is the value of the acceleration of mass $m_1$ that you have mentioned.
Plug it in the equation and you'll find that:
$T_c=\frac{4m_1m_2}{m_1+m_2}{g}$
Let
\begin{align}
m_1 = 12\,\mathrm{kg}, \qquad m_2 = 4\,\mathrm{kg}, \qquad m_3 = 8\,\mathrm{kg}
\end{align}
If you solve this problem symbolically, then you'll find that the tension $T$ applied to mass $m_1$ satisfies
\begin{align}
T = \left(\frac{8m_1m_2m_3}{m_1m_2+m_1m_3+4m_2m_3}\right)g.
\end{align}
If you plug in the values given for the various masses, then you obtain
\begin{align}
T=\frac{192}{17}g \approx (11.30\,\mathrm{kg})g <\text{weight of mass $m_1$},
\end{align}
so it seems that your claim
the tension would be greater than the greatest force of gravity of any mass
is false. For reference, here are the equations you obtain using Newton's Second Law:
\begin{align}
T-m_1g&=m_1a_1\\
\frac{T}{2}-m_2g &= m_2a_2\\
\frac{T}{2}-m_3g&=m_3a_3,
\end{align}
and the constraint you wrote down is correct;
\begin{align}
a_1 = -\frac{1}{2}(a_2+a_3).
\end{align}
Best Answer
No!
As we observe from the figure. The lower pulley has $4kg$ and $8kg$ masses hanging on either side. Since the masses are not equal. The masses will have acceleration and thus the lower pulley has acceleration which accelerates the mass $12kg$.