The "binding energy" is the energy that has to be supplied to disassemble a bound system into its constituent parts. When thermal (milli-eV) neutrons capture on hydrogen, each capture releases a 2.2 mega-eV photon. That energy is the binding energy of deuterium, and hitting deuterium with more than 2.2 MeV allows it to dissociate into free nucleons again.
The nucleon is the ground state of QCD: you can't disassemble a nucleon into lower-mass constituents. (If you hit a nucleon hard enough to make QCD things happen, you can produce a heavier baryon. It is sometimes helpful to imagine the $\Delta$ baryons as bound states of nucleons and pions — but other times that picture is not useful.) So the most reasonable value for the "binding energy" of a free nucleon is zero, just like the "binding energy" of a free electron is zero.
While the free neutron does decay to a proton, an electron, and an antineutrino, the neutron is never a bound state of its weak decay products. Binding energy is the wrong tool to describe that problem. The free neutron and the free proton both have different, nonzero mass excess (a.k.a. mass defect), which is closely related to the binding energy in nuclei with $A>1$.