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
Shell models of the nucleus propose orbitals for nucleons much like those we see for electrons in an atom, but the "distances between them" (measured from regions of maximal likelihood or somesuch) are comparable to the size of the nucleons, so they can't really be thought of as separated in the way one can figure that atomic electrons have regions of maximal likelihood that are distinct from one another. That's why the liquid drop models that nate mentions were pretty useful.
The models require as inputs some potentials and form-factors that are not as well understood as those that affect electrons. Further the shell model must be solved numerically in all but the simplest cases so the results are only approximate.
My dissertation work included taking pictures of the shell distributions (of protons only) in momentum space. Here is one view of a carbon nucleus (after considerable manipulation to approximately remove the final state effects):
the horizontal scale is basically binding energy and you can see the p- (high peak on the left) and s-shells (low peak on the right). The vertical scale has arbitrary units. The lines are to guides the eye only as they have limited physics content.
This figure shows both the energy dependence and the momentum dependence
where the red diamond markers represent the (manipulated to approximately remove final-state interactions) data and the blue circular markers the result of a Monte Carlo calculation based on a shell model using earlier spectral functions and the deĀ Forest prescription for the nuclear form-factors. As before the vertical scale is arbitrary and the horizontal scale is in GeV (natural units).
The double-bump in the momentum data are (in effect) the lobes of the p-orbitals, and the fact that the center is distinctly non-zero is the mark of the s-orbital.
Spatial distributions of bound objects are related to the momentum distributions by a Fourier transformation. You can't actually work that transformation on this figure because it is in energy rather than momentum space, but you might well guess that there would be a lot of spatial overlap and you would be right.