I understand that molecular oxygen has an ionization potential of around 13.6 eV and molecular nitrogen sits at around 14.8 eV; these correspond to wavelengths of around 80 nm and 92 nm, respectively, via $\lambda=hc/I_p$ for $hc=1240\mathrm{\:nm\:eV}$.
My question is how do people manage to ionize with light of much longer wavelengths (example, example, example) if the photon energy for that light is so low? Or are my numbers for the ionization potential of air molecules wrong?
Your conversions are correct - if you want to ionize air molecules with a single photon, then you'll need photons with wavelength shorter than 80nm or so.
However, for the videos you give as examples, that's not the physical mechanism at play. Optical breakdown in air, as in general dielectrics, is an avalanche process. In this process you start with a seed electron, which gets accelerated by the laser field to higher and higher energies until it hits a neutral molecule and, via collisional ionization, knocks out a second electron. You now have two electrons, which get accelerated and hit other molecules, knocking out even more free charge, in an exponential process, which quickly develops into an ionized plasma that can absorb radiation directly, heating up in the process. The emitted light comes from the plasma; the sound comes from its expansion in air and the subsequent collapse of the evacuated bubble it generates.
Now, to have that avalanche process, you need to get that seed electron from somewhere, and here the dynamics change depending on the pulse duration: the seed electrons normally come from a mixture of free electrons naturally present in air (though see also doi / eprint), and electrons ionized via multiphoton ionization, with the relative importance of multiphoton ionization increasing for shorter pulses that get closer to the femtosecond regime.
Now, even if you take the avalanche process out of the equation, a laser with photon energy that's smaller than the ionization potential can still, if it is intense enough, produce ionization via two-photon processes (or higher). This dates all the way back to 1965 [reference, reference] ─ you do need a laser to get the required intensities, but once lasers were invented it took only five years for experiments to show clear indications of six-photon ionization experiments.
Now, the higher the number of photons you need to ionize, the higher the required intensity, so if you want to meaningfully ionize air with laser light in the 1µm range then you're looking at a ~12-photon process, which explains the need for Q-switched pulsed lasers at the very least. In addition to that, the high photon count of the process means that it is highly nonlinear (though still a leading-order perturbative process unlike, say, HHG), i.e. the probability of an $n$-photon process scales as $\mathrm{intensity}^n$, and that explains why the optical breakdown is strongly localized to the optical focus instead of slowly ramping up as the intensity increases.
Also, depending on conditions, if the laser is intense enough then the induced third-order nonlinearity in air can cause an effect called Kerr self-focusing, which is a runaway positive feedback loop where an intense laser induces a "virtual lens" in the air through nonlinear processes, causing it to focus into a tighter beam with a higher intensity, and so on and on until you reach the threshold for optical breakdown. That then means that you don't need such crazy-intense lasers, or particularly clean focusing conditions, to reach those intensity regimes. (For more on what can happen after the breakdown, see laser filamentation.)
And also, if a twelve-photon process sounds crazy, then consider e.g. this paper, with ~150-photon processes, or this one, with processes of order 5000 and higher.
