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.
The final stage of nucleosynthesis at the core of a massive star involves the production
of iron-peak elements, mostly determined by competition between alpha capture and photodisintegration. The starting material is mostly Si28 and weak processes are unable to significantly alter the n/p ratio from unity on short enough timescales. Thus the expected outcome of these quasi-equilibrium reactions should be nuclei with $Z \simeq N$.
Subject to that constraint, then the most stable nucleus resulting from alpha captures onto Si is Ni56.
To produce heavier nuclei (e.g. Zn60) by alpha capture requires higher temperatures (because of the higher Coulomb barrier) and at these higher temperatures photodisintegration drives the equilibrium back towards smaller nuclei.
So where does all the confusion arise? Most of the iron-peak material ejected in a supernova is formed slightly further out from the core in explosive Si burning. The major product is Ni56, as above, and this then undergoes weak decays to Co56 and then Fe56 with half lives of 6 days and 77 days respectively. Thus the most common iron-peak product that ends up in the interstellar medium is Fe56 (also from alpha capture in type Ia supernovae).
Best Answer
The Sun is fabulously inefficient as a fusor: in a given year it consumes only about $10^{-10}$ of its fuel. For the Sun that’s actually a good thing, because it took us five billion years to notice. But an Earthside reactor where you have to confine and control a billion times more fuel than you actually use is quite different from the approach we’ve taken through history.
The Sun also operates at very low power density: in the hottest part of the core, about a hundred watts per cubic meter. You can imagine reproducing this power density by filling the inside of your house with scaffolding and operating one light bulb every cubic meter. It’d be warm, for sure, but not solar-warm; you might not even need to re-wire your house. You could get higher power density than the core of the Sun just by setting your house on fire.
The reason the Sun is still useful as a power source, despite its fabulous inefficiency, its low power density, and its astronomical distance, is that the Sun’s power-generating core is mind-bogglingly large.
At constant power density, the amount of effort it takes to confine a hot plasma goes roughly like its surface area (because you only have to push on the outside surface). Suppose I have a tokamak which produces a certain amount of fusion power and has a certain cost to keep the plasma contained. If I build the same design but three times bigger across, I expect the cost of confining the plasma to increase by a factor $3^2 = 9$. But if I keep the energy density the same, the energy output should go up like the volume, by a factor of $3^3 = 27$. The engineering question then becomes whether I can make the tokamak large enough that the power produced outstrips the containment cost, while still having it fit on Earth, in a building. (A big building would be okay.)