The analysis of the phase structure of gauge theories is a whole field. Some major breakthroughs were the t'Hooft anomaly matching conditions, the Banks-Zaks theories, Seiberg duality, and Seiberg Witten theory. There is a lot of controversy here, because we don't have experiment or simulation data for most of the space, and there is much more unknown than known.
The first thing to note is that when the Higgs field vacuum expectation value is zero, the Higgs doesn't touch the low energy physics. You can ignore the Higgs at energy scales lower than it's mass, and if this mass is much greater than the proton mass, the result is indistinguishable qualitatively from the Higgsless standard model. So I'll describe the Higgsless standard model.
Higgsless standard model
Even without the Higgs, electroweak symmetry is broken anyway by QCD condensates. When the Higgs VEV is zero, the W and Z do not become completely massless, although they become much much lighter.
The reason is that QCD has a nontrivial vacuum, where quarks antiquark pairs form a q-qbar scalar fluid that breaks the chiral symmetry of the quark fields spontaneously. This phenomenon is robust to the number of light quark flavors, assuming that there aren't so many that you deconfine QCD. QCD is still asymptotically free with 6 flavors, and it should be confining even with 6 flavors of quarks. So I have no compunctions about assuming the confinement mechanism still works with 6 flavors, and all 6 are now like the up and down quark. Assuming the qualitative vacuum structure is analogous to QCD is plausible and consistent with the anomaly conditions, but if someone were to say "no, the vacuum structure of QCD with 6 light quarks is radically different from the vacuum structure of QCD", I wouldn't know that this is wrong with certainty, although it would be strange.
Anyway, assuming that QCD with 6 light quarks produces the same sorts of condensates as QCD with 3 light quarks (actually 2 light quarks and a semi-light strange quark), the vacuum will be filled with a fluid which breaks SU(6)xSU(6) chiral rotations of quark fields into the diagonal SU(6) subgroup. The SU(6) is exact in the strong interactions and mass terms, it is only broken by electroweak interactions.
The electroweak interactions are entirely symmetric between the 3 families, so there is a completely exact SU(3) unbroken to all orders. The SU(6)xSU(6) breaking makes a collection of massless Goldstone bosons, massless pions. The number of massless pions is the number of generators of SU(6), which is 35. Of these, 8 are exactly massless, while the rest get small masses from electroweak interactions (but 3 of the remaining 27 go away into W's and Z's by Higgs mechanism, see below). The 8 massless scalars give long-range nuclear forces, which are an attractive inverse square force between nuclei, in addition to gravity.
The hadrons are all nearly exactly symmetric under flavor SU(6) isospin, and exactly symmetric under the SU(3) subgroup. All the strongly interacting particles fall into representation of SU(6) now, and the mass-breaking is by terms which are classified by the embedding of SU(3) into SU(6) defined by rotating pairs of coordinates together into each other.
The pions and the nucleons are stable, the pion stability is ensured by being massless, the nucleon stability by approximate baryon number conservation. At least the lowest energy SU(3) multiplet
The condensate order-parameter involved in breaking the chiral SU(6) symmetry of the quarks is $\sum_i \bar{q}_i q_i$ for $q_i$ an indexed list of the quark fields u,d,c,s,t,b. The order parameter is just like a mass term for the quarks, and I have already diagonalized this order parameter to find the mass states. The important thing about this condensate is that the SU(2) gauge group acts only on the left-handed part of the quark fields, and the left-handed and right handed parts have different U(1) charge. So the condensate breaks the SU(2)xU(1) gauge symmetry.
The breaking preserves a certain unbroken U(1) subgroup, which you find by acting the SU(2) and U(1) generators. The left handed quark field has charge 1/6 and makes a doublet, so for the combination $I_3+Y/2$ where I is the SU(2) generator and Y is the U(1) generator, you get a transformation of 2/3 and 1/3 on the top and bottom component, which is exactly the same as $I_z + Y/2$ on the singlets (since they have no I). So this combination isn't chiral, and preserves the vacuum. So the QCD vacuum preserves the ordinary electromagnetic subgroup, which means it makes a Higgs, just like the real Higgs, which breaks the SU(2)xU(1) down to U(1) electromagnetic, with W and Z bosons just like in the standard model.
This is not really as much of a coincidence as it appears to be--- a large part of this is due to the fact that QCD condensates in our universe are not charged, so that they don't break electromagnetism, because u-bar and u have opposite electromagnetic charge transformation. This means that a u-bar u condensation leaves electromagnetism unbroken, and it isn't a surprise that it doesn't leave any of the rest of SU(2) and U(1) unbroken, because it's a chiral condensate, and these are chiral gauge transformations.
The major difference is that there are 3 separate Higgs-like condensates, one for each family, each with an identical VEV, all completely symmetric with each other under the global exact SU(3) family symmetry.
The W's and Z's get a mass from an arbitrary one of these 3, leaving 2 dynamical Higgs-like condensates. The main difference is that these scalar condensates don't necessarily have a simple distinguishable higgs-boson-like oscillation, unlike a fundamental scalar Higgs. The result of this is that the W's and Z's acquire QCD-scale masses, so around 100 MeV for the W's and Z's, as opposed to approximately 100 GeV in the real world. The ratio of the W and Z mass is exactly as in the standard model.
Behavior of analogs of ordinary objects
The low energy spectrum of QCD is modified drastically, due to the large quark number. The 8 massless pions and 24 nearly massless pions (three of the pions are eaten by the W's and Z's to become part of the massive vectors) include all the diquark degrees of freedom that we call the pions,kaons and certain heavy quark mesons. There will still be a single instanton heavy eta-prime from the instanton violated chiral U(1) part of U(6)xU(6). There should be 35 rho particles splitting into 8 and 27 and 35 A particles splitting into 8 and 27 effectively gauging the flavor symmetry.
The 6 quarks could be thought of as getting a mass from their strong interaction with the Higgs-like condensates, of order some meVs, but since the mass of a quark is defined at short distances, from the propagator, it might be more correct to say the quarks are massless. Some of the particles you see in the data-book, the sigma(660), the f0(980) should disappear (as these are weird--- they might be the product of pion interactions making some extremely unstable bound states, something which wouldn't work with massless pions)
The electron and neutrino will be massless except for nonrenormalizable quark-lepton direct coupling, which would couple the electron to the Higgslike chiral quark condensate. This effect is dimension 6, so the compton wavelength of the electron will be comparable to the current radius of the visible universe. The neutrino mass will be even more strongly suppressed, so it might as well be exactly massless.
The massless electron will lead the electromagnetic coupling (the unHiggsed U(1) left over below the QCD scale) to logarithmically go to zero at large distances, from the log-running of QED screening. So electromagnetism, although it will be the same subgroup of SU(2) and U(1) as in the Higgsed standard model, will be much weaker at macroscopic distances than it is in our universe.
Nuclei should form as usual at short distances, although Isospin is now a nearly exact SU(6) symmetry broken only by electromagnetism, and not by quark mass, and with an exact SU(3) subgroup. So all nuclei come in SU(6) multiplets slightly split into SU(3) multiplets. The strong force will be longer ranged, and without the log-falloff of the electromagnetic force, because the pions quickly run to a free-field theory, since the pion self-interactions are of a sigma-model type. The pion interactions will look similar to gravity in a Newtonian approximation, but scalar mediated, so not obeying the equivalence principle, and disappearing in scattering at velocities comparable to the speed of light.
The combination of a long-ranged attractive nuclear force and a log-running screened electromagnetic force might give you nuclear bound galaxies, held at fixed densities by the residual slowly screened electrostatic repulsion. These galaxies will be penetrated by a cloud of massless electrons and positrons constantly pair-producing from the vacuum.
Best Answer
Theoretically, yes it should be possible to derive the boiling point of diatomic nitrogen from fundamental forces. In fact, you don't even need to involve the strong force or weak force (or the strong nuclear force, which is sort of different). The strong forces bind the quarks together into nucleons and the nucleons together into nuclei, but they have essentially no effect on distance scales much larger than that of an atomic nucleus. So, for purposes of calculating the boiling point of nitrogen, you can treat the nucleus as basically a point charge. The only force that is relevant to calculating a boiling point is the electromagnetic force.
Now the bad news: even something as simple as calculating the energy levels of helium, with 2 electrons, is impossible to do analytically. To analyze the behavior of even just those two electrons (and He nucleus) in detail, you need to use either perturbation theory or a numerical simulation, or both. And of course, the complexity increases with the number of particles, so simulating the 14 electrons and 2 nuclei of a nitrogen molecule is absurdly complicated. Perhaps it's been done, but I'm not a condensed matter physicist so I wouldn't know where to look for a reference. Maybe someone else can provide you with that information.
If you were to calculate the boiling point of nitrogen, I believe the main effect that you'd take into account would be the instantaneous dipole interaction. According to the Wikipedia article, it gives an interaction energy in terms of the polarizabilities and ionization energies of the molecules. Those are the quantities that you would have to extract from your simulation and/or perturbative calculation of the dynamics of the nitrogen molecule, if you wanted to calculate the effect from first principles.
$$E_{AB}^{\rm disp} \approx -{3 \alpha^A \alpha^B I_A I_B\over 4(I_A + I_B)} R^{-6}$$
(that formula is actually for monatomic noble gases, it may not apply to diatomic molecules)
Once you get the interaction energy as a function of intermolecular separation $R$, you would then have to do either another numerical simulation, or a rather complicated calculation, to show that a large pool of nitrogen molecules subject to the given intermolecular force undergoes a phase transition at 77.36K (at standard pressure, I assume). There are various thermodynamic models you could use, some more accurate than others, but of course the more accuracy you want, the more computation power you'll need. I suspect that in order to get within a few degrees of the actual temperature, you would need to do something more computationally intensive than would be possible by hand.