The neutrino mass is less than one eV --- that is, less than $0.000\,001\rm\,MeV$. So leaving its mass out of the calculation isn't a big deal.
The energy is carried away by all three reaction products. Consider a case where the initial momentum of the system is zero, and the geometry of the reaction products means that they all three have the same momentum magnitude $p$. (Perhaps they all make 120 degree angles with each other's paths.) Then the kinetic energy carried by the (ultra-relativistic) neutrino is $K_\nu = pc$, the kinetic energy carried by the non-relativistic deuterium nucleus is $K_{\rm d} = p^2/2M_{\rm d}$, and the positron's kinetic energy
$$
K_{\rm e}=
\frac{p^2}{(\gamma+1)m_{\rm e}}
=(\gamma-1)m_{\rm e}c^2
,
$$
with the usual relativistic factor $\gamma=(1-v^2/c^2)^{-1/2}$,
doesn't usually fit into either of those nice approximation regions.
But you can see that the deuteron, with the most mass, carries off the smallest amount of kinetic energy in that (and actually all) geometries.
How the energy is shared among the light particles as the geometry of the interaction changes is a fun problem to play with.