How are octet-rule models supported by thermodynamics and quantum theory?
You can't really compare the energy of $\ce{Cl-}$ to $\ce{Cl.}$, since the two states have different composition. Instead, you're interested in the difference in energy between $\ce{Cl-}$ and $\ce{Cl. + e-}$, where the electron is completely free and resting in a vacuum with no interactions with anything else. We give that electron a reference potential energy of 0. Then the question becomes, how much potential energy is lost by interacting with various atoms, and the thermodynamically favorable state is the one with the most negative potential energy.

If we consider nuclei to be point positive charges and electrons to be negative charge distributed in a much wider space around the nucleus, we find that as that free electron gets closer to the nucleus, it experiences an electrostatic interaction with the positively charged nucleus even if the atom overall is neutral. We can simplistically estimate the magnitude of the potential energy loss based on basic principles of electrostatics, where potential energy is a function of $\frac{q^+q^-}{r}$. 

Where molecular orbital theory comes in is when we calculate the charge and the distance r. Simplistically, orbitals that are higher in energy on the same atom are higher in energy because they have a longer average distance from the nucleus (denominator in PE term is bigger) and because there is more electron density from other electrons between them and the nucleus, so the "effective" positive charge of the nucleus is reduced (numerator in PE is smaller). This is called "shielding". 

Slater's rules are simple heuristics that provide a fairly good approximation of the effective nuclear charge experienced by an electron based on the atomic number and orbital. 

Conceptually, then, the answer to your question is that adding an electron to a neutral atom like $\ce{Cl.}$ is most favorable when the electron experiences the best combination of effective nuclear charge and short average distance to the nucleus. This happens when adding the last electron to an octet because that is the element of that period with the greatest nuclear charge. Adding an electron to the next neutral element Argon is much less favorable because it has to go in the next shell (n=4) meaning that it has a larger r (larger denominator) and has much more electron density between it and the nucleus, so the effective nuclear charge is less (smaller numerator) even though the atomic number of Ar is one more than Cl.

For $\ce{Na+}$, it isn't correct to say that the system has lower energy after an electron is removed, but rather that it takes a lot less energy to remove that electron than an electron from neutral neon, which hopefully makes sense based on the discussion above.