Energy of a fission nuclear bomb comes from the gravitational energy of the stars.
Protons and neutrons can coalesce into different kinds of bound states. We call these states atomic nuclei. The ones with the same number of protons are called isotopes, the ones with different number are nuclei of atoms of different kinds.
There are many possible different stable states (that is, stable nuclei), with different number of nucleons and different binding energies. However there are also some general tendencies for the specific binding energy per one nucleon (proton or neutron) in the nuclei. States of simple nuclei (like hidrogen or helium) have the lowest specific nucleon binding energy amongst all elements, but the higher is the atomic number, the higher the specific energy gets. However, for the very heavy nuclei the specific binding energy starts to drop again.
Here is a graph that sums it up:
http://en.wikipedia.org/wiki/File:Binding_energy_curve_-_common_isotopes.svg
It means that when nucleons are in the medium-atomic number nuclei, they have the highest possible binding energy. When they sit in very light elements (hidrogen) or very heavy ones (uranium), they have weaker binding. Thus, one can say that for the low "every-day" temperatures, the very heavy elements (like the very light ones) are quasistable in a sense.
Fission bomb effectively "lets" the very heavy atomic nuclei (plutonium, or uranium) to resettle to the atoms with lower number of nucleons, that is, with higher bound energies. The released binding energy difference makes the notorious effect. In terms of the graph cited above, it corresponds to nucleons moving from the right end closer to the peak.
Yet this is not the only way to let nucleons switch to the higher binding energy state than the initial one. We can "resettle" very light elements (like hydrogen) and let nucleons move to the peak from the left. That would be fusion.
Heavy nucleons emerge in the stars. Here the gravitational energy is high enough to let the nucleons "unite" into whatever nuclei they like. Stars usually are formed from the very light elements and the nucleons inside, again, tend to get to the states with lower energies, and form more "medium-number" nuclei. The energy difference powers stars and we see the light emission, high temperatures and all other fun effects.
However, sometimes the temperatures in the stars are so high, that nucleons form the very heavy nuclei from the medium-number nuclei. even though there is no immediate "energy" benefit.
These heavy elements then disseminate everywhere with the death of the star. This stored star energy can then be released in the fission bomb.
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.
Best Answer
The answer lies in this diagram
It shows the binding energy per nucleon over the number of nucleons n in a nucleus. As you can see, it has a maximum around Iron (n=56), which means that left from iron energy can be released by fusion. For elements heavier than iron you need to put in energy to fuse them together, or equivalantly that fission is energetically favorable. That is why conditions like in a supernova are necessary to produce the heavier elements.
So, fusioning ${}^2H$ to ${}^4He$ releases energy, but the reverse fission process costs energy. There is no net gain, even in a perfect aparatus.