Firstly, the diagonal action of $G$ on $\mathcal{O}_1\times\mathcal{O}_2\times\mathcal{O}_3$ (I'm altering your notation for later convenience) is not necessarily free: if there are any points $(\alpha_1,\alpha_2,\alpha_3)\in\mathcal{O}_1\times\mathcal{O}_2\times\mathcal{O}_3$ with $\mathfrak{g} _{\alpha_1}\cap\mathfrak{g} _{\alpha_2}\cap\mathfrak{g} _{\alpha_3}\ne\lbrace0\rbrace$, then $G$ will have nontrivial isotropy group there. So it's not clear to me that $(\mathcal{O}_1\times\mathcal{O}_2\times\mathcal{O}_3) / G$ is even a differentiable manifold. Supposing the freeness condition is met (e.g. the $\alpha_i$ are nonzero on different simple blocks of a semisimple Lie algebra), the value of the reduced form is obtained from the lifted expression
$$
\sum_{i=1}^3\ \omega_{KKS}(\alpha_i)(-\rm{ad} _{\xi_i}^*\alpha_i, -\rm{ad} _{\zeta_i}^*\alpha_i) = \sum_{i=1}^3\alpha_i([\xi_i,\zeta_i])
$$
where $\sum_{i=1}^3\alpha_i=0$ and $\sum_{i=1}^3(-\rm{ad} _{\xi_i}^*\alpha_i)=\sum_{i=1}^3(-\rm{ad} _{\zeta_i}^*\alpha_i)=0$ (expressing the restriction to $\mu^{-1}(0)$). However, by the general theory of reduction, the $\mathcal{O}_1\times\mathcal{O}_2\times\mathcal{O}_3$ symplectic form, when restricted to $\mu^{-1}(0)$, is degenerate along directions

$$(-\rm{ad} _\eta^*\alpha_1,-\rm{ad} _\eta^*\alpha_2,-\rm{ad} _\eta^*\alpha_3),\qquad\eta\in\mathfrak{g}.$$
So subtract such a vector with $\eta=\xi_2$ from the first slot and $\eta=\zeta_3$ from the second. The expression reduces to
$$
\alpha_1([\xi_1-\xi_2,\zeta_1-\zeta_3]).
$$

That's the simplest form I can see right now. It would probably help if you provided some context for your question. Why do you want to know this anyway?