To understand binding energy and mass defects in nuclei, it helps to understand where the mass of the proton comes from.
The news about the recent Higgs discovery emphasizes that the Higgs mechanism gives mass to elementary particles. This is true for electrons and for quarks which are elementary particles (as far as we now know), but it is not true for protons or neutrons or for nuclei. For example, a proton has a mass of approximately $938 \frac{\mathrm{MeV}}{c^2}$, of which the rest mass of its three valence quarks only contributes about $11\frac{\mathrm{MeV}}{c^2}$; much of the remainder can be attributed to the gluons' quantum chromodynamics binding energy. (The gluons themselves have zero rest mass.) So most of the "energy" from the rest mass energy of the universe is actually binding energy of the quarks inside nucleons.
When nucleons bind together to create nuclei it is the "leakage" of this quark/gluon binding energy between the nucleons that determines the overall binding energy of the nucleus. As you state, the electrical repulsion between the protons will tend to decrease this binding energy.
So, I don't think that it is possible to come up with a simple geometrical model to explain the binding energy of nuclei the way you are attempting with your $\left(1\right)$ through $\left(15\right)$ rules. For example, your rules do not account for the varying ratios of neutrons to protons in atomic nuclei. It is possible to have the same total number of nucleons as $\sideset{^{56}}{}{\text{Fe}}$ and the binding energies will be quite different the further you move away from $\sideset{^{56}}{}{\text{Fe}}$ and the more unstable the isotope will be.
To really understand the binding energy of nuclei it would be necessary to fully solve the many body quantum mechanical nucleus problem. This cannot be done exactly but it can be approached through many approximate and numerical calculations. In the 1930's, Bohr did come up with the Liquid Drop model that can give approximations to the binding energy of nuclei, but it does fail to account for the binding energies at the magic numbers where quantum mechanical filled shells make a significant difference. However, the simple model you are talking about will be incapable of making meaningful predictions.
EDIT: The original poster clarified that the sign of the binding energy seems to be confusing. Hopefully this picture will help:
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This graph shows how the potential energy of the neutron and proton that makes up a deuterium nucleus varies as the distance between the neutron and proton changes. The zero value on the vertical axis represents the potential energy when the neutron and proton are far from each other. So when the neutron and proton are bound in a deuteron, the average potential energy will be negative which is why the binding energy per nucleon is a negative number - that is we can get fusion energy by taking the separate neutron and proton and combining them into a deuteron. Note that the binding energy per nucleon of deuterium is $-1.1 \, \mathrm{MeV}$ and how that fits comfortably in the dip of this potential energy curve.
The statement that $\sideset{^{56}}{}{\text{Fe}}$ has the highest binding energy per nucleon means that lighter nuclei fusing towards $\text{Fe}$ will generate energy and heavier elements fissioning towards $\text{Fe}$ will generate energy because the $\text{Fe}$ ground state has the most negative binding energy per nucleon. Hope that makes it clear(er).
By the way, this image is from a very helpful article which should also be helpful for understanding this issue.
From Wikipedia:
Iron-56 (56Fe) is the most efficiently bound nucleus meaning that it has the least average mass per nucleon. However, nickel-62 is the most tightly bound nucleus in terms of energy of binding per nucleon. (Nickel-62's higher energy of binding does not translate to a larger mean mass loss than Fe-56, because Ni-62 has a slightly higher ratio of neutrons/protons than does iron-56, and the presence of the heavier neutrons increases nickel-62's average mass per nucleon).
In any case, iron-56 is not necessarily usually shown (anymore) as the highest binding energy per nucleon, mostly because it's not usually possible to show the tiny difference in binding energy between iron-56 and nickel-62. The difference is so tiny that accounting for the fact that the neutron is 0.1% heavier than the proton, instead of treating all nucleons equally, changes which is bound more tightly. Because of this tiny difference, it's unlikely that you'll see one depicted as higher than the other on any plot that actually shows both of them, like this one (source: https://www.asc.ohio-state.edu/kagan.1/phy367/Lectures/P367_lec_14.html):
Can you tell which of the two has higher binding energy? I certainly can't.
In addition, adding neutrons does not always make the nucleus more stable. In the nuclear shell model, protons and neutrons, being distinguishable from each other, occupy distinct sets of energy levels in the nucleus. (As a side note, the neutron energy levels are shifted lower than the proton energy levels because there is no Coulomb repulsion among neutrons. This is why the heavier stable nuclei tend to have more neutrons than protons - you can fill more of the neutron energy levels for the same valence nucleon energy.) Since they're both fermions, you have to add new neutrons to higher and higher energy levels. Adding too many neutrons, enough to put the highest filled neutron energy level significantly above the highest proton energy level, will cause the highest-energy neutrons to spontaneously convert to protons via beta decay so that they can occupy a lower (proton) energy level.
Best Answer
I don't see any theoretical boundaries, like those in the efficiency of thermodynamic machines. The limit will come from what we call "fusion" and what instead we call "other particle interactions".
An electron and an anti-electron, both stable particles, can interact and "fuse", becoming just pure photons, so in this example the fusion reaction electrons -> photons is 100% efficient.
I will call "fusion" a reaction between particles that we can collect and harvest, then get positive net energy from the fusion. Hydrogen, Deuterium and Tritium, we can harvest them from water, then make them fuse to release more energy. Anti-electron, no, there aren't many free anti-electron around that we can practically use. Similar story for "quark fusion": we may be able to observe it in collision experiments, but we are unable to pick free quarks from the ocean (there are some hiding, I know!) and collect enough of them to produce useful energy.
There may be some way to force the decay of protons, so we can transform most of their mass in energy. Anyway, I won't call it fusion. Is the Nickel nucleus really the lowest-energy architecture? We may ask to a neutron star once get frozen.
If we restrict to normal atoms, the most efficient practical fusion known is the production of Helium from hydrogen isotopes; there are probably some other very unstable isotopes that may release more energy. In collision experiments, we can create extremely unstable isotopes that release a lot of energy when recombining to more stable states. Is that fusion? Yes, but not a practical source of energy.
In practical applications with stable isotopes, I think that the "mass excess" of an isotope can be regarded as a measure of the theoretical maximum of energy that can be harvested. Your calculation to get Nickel from bare nucleons is likely the top.
Summary: there is no upper bound in the initial energy of free particle that we can combine. Lower bound (excluding antimatter) maybe the Proton and the Nickel nucleus. Maybe.