Why does the energy of a thermodynamic system depend on the volume?

In thermodynamics one assumes that the energy (or other observables) of the system only depend on a few parameters like the temperature, the particle number and also the volume:

$dE = \left(\frac{\partial E}{\partial T}\right) dT + \left(\frac{\partial E}{\partial N}\right) dN + \left(\frac{\partial E}{\partial V}\right) dV := SdT + \mu dN - pdV$.

Why is the volume enough to describe the whole geometry and details of the container? Why isn't there another term e.g. for the surface? Or some geometry dependence?

On the microscopic level the energy levels depend on the geometry (For example a spherical and a rectangular box have different energy levels.). Also like in the case of a gas trapped in a harmonic oscillator potential $V(x) = k\cdot x^2$ there is no well-defined "volume" of the gas. However if one would increase $k$ one would also increase the energy of the system. So in this case the energy depends on $k$ and not on the volume (Probably there is some way to calculate the characteristic gas "volume" out of $k$, but this would be only a rough estimate.).

Is there any argumentation (from a fundamental point of view) why the energy of a huge system only depends on the volume and one can neglect everything else?

EDIT: As I understood thermodynamics you basically start from the microscopic point of view. There you have a Hamiltonian. Then one imposes a density matrix (micro canonical, canonical, grand canonical). This defines the whole dynamics of the system. Then you take the limit $N \to \infty$ (this allows replacing the sum over all states by the integral). This can be exactly done in the case of a ideal gas in a rectangular container. There you find out that the energy of the system only depends on the temperature $T$, the particle number $N$ and the volume $V$. Calculating the total differential of the energy $E = \left(\frac{\partial E}{\partial T}\right) dT + \left(\frac{\partial E}{\partial N}\right) dN + \left(\frac{\partial E}{\partial V}\right) dV := SdT + \mu dN - pdV$ one can find expressions for $S,\mu$ and $p$ which are defined by this equation.

For all other systems where one cannot solve it exactly one imposes that after taking $N \to \infty$ the energy only depends on $T, N$ and $V$. For me it's completely clear that the energy depends on $N$ (since the Hamiltonian is linear with the particle number) and on $T$ (because it is part of the density matrix).

Apart from that it is also quite clear that there must be a term that represents the change in the potential (i.e. work). According to the first law this work is only given by the change in volume. That is the point I don't understand: why should the change of potential energy only depend on the volume for $N \to \infty$?

Two examples for that:

1. 
   

   There are potentials (like the harmonic oscillator, but many more) where there is no fixed border and so there is no sensible definition of volume. In the case of a harmonic oscillator I can also solve it explicitly and find that the change of the potential energy depends on $\omega_x\omega_y\omega_z$ which one can relate to a "typical volume" $V$ of the harmonic oscillator. Using this volume I can the bring the energy dependence in the same form as in the first law. However for any other potential I would need to guess a "typical volume" $V$ (since I can't solve the equations). Even if I can find a equation that looks like the first law and has a term $\sim dV$ the pressure would depend on my "typical volume" (for example if I choose my "typical volume" to be twice as big then the pressure has only half of its value). This sounds pretty strange to me because if I have a gas trapped in a harmonic potential I can measure the pressure and get some well defined value.

   
   



2. 
   

   My second idea was that maybe one only considers system where the potential is either 0 or infinite. There one has a fixed geometry. But if I don't have a rectangular geometry the energy levels differ. So there is a clear dependence on the geometry. So why in the limit $N \to \infty$ only a term depending on the volume survives? This means that if I change the geometry of my container completely but keep the volume constant nothing changes. I don't see any fundamental reason why this should be the case.