How to explain $E=mc^2$ mass defect in fission/fusion?

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:

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.

Before I begin, though, I wanted to point out that the original poster talks about some of his points being "wrong" because of conflicting info. If you want to have a good discussion about the fundamentals of nature (which is almost purely philosophical, mind you), it is better to abandon absolutes. There is no right or wrong, good or bad. Forget about law and think only theory. Scientists, so far, have mostly agreed that the simplest (i.e., easiest to understand) explanation is best and that nothing is resistant to change or evolution. Though, some scientists in the main stream have been behaving poorly in my opinion. Big money and politics is a poor environment for good science.

Ok, with that out of the way, I'd like to add in some points about $\sideset{^{56}}{}{\text{Fe}}$. One of the comments on the question states:

The prevalence of iron comes in part from the decay Ni-56 → Co-56 → Fe-56. Ni-56 is a preferred byproduct of supernovae in part due to having equal numbers of protons and neutrons (nuclear reactions have kinetics to consider as well as thermodynamics).

I'm unaware of $\sideset{^{56}}{}{\text{Ni}}$ as a major byproduct of supernovae, BUT I am familiar with stellar evolution theory which states that when a star begins to burn (i.e., fusion) $\sideset{^{56}}{}{\text{Fe}}$ in its core, the sudden reduction in energy output related to the lower per nucleon binding energy of the product from fusing $\sideset{^{56}}{}{\text{Fe}}$ results in structural collapse of the star and hence the supernovae. Here's a source for more info: "Q: Why does iron kill stars?".

Now, why does $\sideset{^{56}}{}{\text{Fe}}$ have the "strongest" binding energy per nucleon? Well, I asked that question to a very knowledgeable professor in my undergraduate days and the answer was simple and elegant. The physical diameter of the $\sideset{^{56}}{}{\text{Fe}}$ nucleus is approximately equal to the distance over which the strong force can act before EM begins to take over. This means that every nucleon (proton, neutron) in the $\sideset{^{56}}{}{\text{Fe}}$ "feels" maximum pull via strong force due to no one nucleon being "out of range" of another nucleon's pull. This idea of a "stuffed" spherical arrangement of nucleons in combination with the idea that the strong force is dependent on neutron/proton proportions (which $\sideset{^{56}}{}{\text{Fe}}$ has the "super magic number" or whatever it is called) creates a situation where every nucleon in a $\sideset{^{56}}{}{\text{Fe}}$ nucleus reaches "maximum" binding energy (i.e., pull from all other nucleons) according to theory.