Let us look at a single geometric fiber $X$. Let $L^\delta(\bar\omega) L^\Delta(\bar\omega)|_X = \mathcal{O}_X(D)$. The Riemann-Roch theorem for abelian varieties (Mumford "Abelian Varieties", Chap. 3 Section 16) states that $$\chi(\mathcal{O}_X(D)) = D^g/g!$$ and moreover that $\chi(\mathcal{O}_X(D))^2 = \deg \phi$, where $\phi$ is the polarization map defined by $\mathcal{O}_X(D)$. So the Hilbert polynomial $\chi(\mathcal{O}_X(3nD)) = 3^gn^gD^g/g! = 3^g n^g \chi(\mathcal{O}_X(D)) = 3^g n^g d$. I got almost the right answer (where did $2^g$ go?), so maybe I misunderstood the question, but maybe I hope this is still helpful.
EDIT. Is the superscript $\Delta$ the symmetrization of the line bundle in question? Then it would explain why the $2^g$ above is missing...
Let us look at a single geometric fiber $X$. Let $L^\delta(\bar\omega) = \mathcal{O}_X(D)$. The Riemann-Roch theorem for abelian varieties (Mumford "Abelian Varieties", Chap. 3 Section 16) states that $$\chi(\mathcal{O}_X(D)) = D^g/g!$$ and moreover that $\chi(\mathcal{O}_X(D))^2 = \deg \phi$, where $\phi$ is the polarization map defined by $\mathcal{O}_X(D)$. So the Hilbert polynomial $\chi(\mathcal{O}_X(3nD)) = 3^gn^gD^g/g! = 3^g n^g \chi(\mathcal{O}_X(D)) = 3^g n^g d$. I got almost the right answer (where did $2^g$ go?), so maybe I misunderstood the question, but maybe this is still helpful.
EDIT. Is the superscript $\Delta$ the symmetrization of the line bundle in question? Then it would explain why the $2^g$ above is missing...