Most of the universe's nitrogen is formed in larger, main sequence stars using the CNO Cycle, right?
But I cannot find a good, specific explanation as to why $^{14}$N, with both an odd number of neutrons and protons, is formed preferentially to $^{15}$N?
Both $^{14}$N and $^{15}$N are produced as part of the CNO cycle during the hydrogen-burning main sequence phase of stars more massive than the Sun. However $^{15}$N reacts rapidly with protons to (re)form $^{12}$C and an alpha particle, whereas the much slower $^{14}$N$(p,\gamma){}^{15}$O reaction allows $^{14}$N abundances to build up and dominate when the CNO cycle reaches an equilibrium.
Details:
The addition of protons to $^{14}$N $$ p + {}^{14}{\rm N} \rightarrow {}^{15}{\rm O} + \gamma$$ is the slowest reaction in the CNO cycle and hence at equilibrium there is a build up of $^{14}$N.
As to why this is the slowest reaction in the cycle; it is likely because:
(i) Of the other proton addition reactions in the cycle, adding a proton to a carbon nucleus has a lower Coulomb barrier so is faster.
(ii) The beta decay reactions, although governed by the weak interaction, are not subject to the high Coulomb barriers of the proton addition reactions and so they are faster.
(iii) That leaves $$ p + {}^{15}{\rm N} \rightarrow {}^{12}{\rm C} + \alpha$$ which is a faster reaction than adding a proton to $^{14}$N because whilst the latter is a "radiative capture" reaction involving an electromagnetic transition resulting in a gamma ray (see Brune & Davids 2015), the former is a more rapid (by 4 orders of magnitude) strong force interaction. This will also be the reason why $^{15}{\rm N}(p,\alpha){}^{12}$C is totally dominant over $^{15}{\rm N}(p,\gamma){}^{16}$O and as a result allows the $^{12}$C to be regenerated; and means that there is a CNO cycle at all!