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}