Why is classical electromagnetism linear? When I ask this, I mean it as in when a test charge $q$ is placed in a region that contains two fixed charges $q_1$ and $q_2$, the force acting on it is the vector sum of the forces it would experience when placed alone with either of the two charges. In essence, I'm asking about the principle of superposition. Why is it that charges don't have effects on each other's fields? Is there some sort of symmetry argument that can be used to show that?
I'm also trying not to simply accept the argument that Maxwell's equations are linear, as from my understanding their linearity is more postulated than anything else.
EDIT: So many quantum explanations! Those are all fine and good, but I was hoping for something that did not have to rely on more advanced and fundamental particle physics, just a classical argument.
 A: From a Classical perspective, this is because (as you mention) Maxwell's equations are linear. That is to say that the theory is linear (an abstract mathematical property) because it is represented by an abstract mathematical formalism which has that property.
That said, perhaps you will find this more satisfying: Photons have no charge. We know that the electromagnetic force is delivered via photons, so if the fields of two charges were to interfere with one another, then that means that photons are interacting with and disrupting the behavior of other photons. But only particles with charge (i.e. "charges") experience the electromagnetic force. Photons carry no charge; ergo, photons do not affect the behavior of other photons.
A: Electromagnetism is a gauge theory with the gauge group U(1), which is abelian. This means there are no terms in the EM Lagrangian where the electromagnetic field interacts with itself. 
$L = \frac{-1}{4}F^2 + \bar{\Psi}(i\gamma^{\mu}D_{\mu} - m)\Psi$, where D is the covariant derivative containing the gauge field $A_{\mu}$, at tree (classical) level there are no photon-photon interactions. This carries over to the classical theory.
Of course there are higher order effects with photons interacting with each other via. loops of fermions, but this is no longer classical. 
Hope this is what you were looking for.
A: The Newtonian limit of general relativity is also linear, even though there is nothing inherently linear about general relativity. Why should it be?
Because most physical laws are linear when the input is close enough to zero. For example, we can model thermal expansion as linear because if we expand $V(T)$ as a Taylor series $V(T + \Delta T) = V(T) + V'(T) \Delta T + \frac{1}{2}V''(T) (\Delta T)^2 + \ldots$ and $\Delta T$ is small enough, then we can neglect all higher order terms.
We can say the same thing about classical electromagnetism. We should expect the electric field intensity at a point $E(x)$ to be expressed as some Taylor series in functional derivatives, $E(x) = E_0 + \int \rho(x') \frac{\delta E(x)}{\delta \rho(x')} \, \mathrm{d}^3x' + \frac{1}{2} \iint \rho(x') \rho(x'') \frac{\delta^2 E(x)}{\delta \rho(x') \delta \rho(x'')} \, \mathrm{d}^3 x' \, \mathrm{d}^3 x'' + \ldots$, where if $\rho$ is small enough then we can neglect higher-order terms and think of $E$ as linear in $\rho$. (Here $E_0$ comes from the boundary conditions.)
We simply happen to find ourselves at a point in time where the universe has cooled enough so that the fields around us are low enough to fall within the linear regime. A fraction of a second after the Big Bang, this was not the case and classical electrodynamics would not have been an accurate description of the electromagnetic interaction.
A: It is enough to show one counterexample to see that the observed and claimed linearity of electrostatics is a simplification. Your example is

a test charge q is placed in a region that contains two fixed charges q1 and q2, the force acting on it is the vector sum of the forces it would experience when placed alone with either of the two charges.

Let’s move the charge q1 behind q2, so that the three charges are aligned:

In my understanding the charge q1 is now perfectly shielded by the charge q2 and the test charge feel only charge q2.
To round up the answer I dare say that due to the mutual displacement of the fields from q1 and q2 the force between q2 and the test charge should be a little bit stronger than without q1, but of course not the vector sum from q1 and q2 with the test charge.

Since there is a downvote perhaps I have to add a little bit more explanation.
Having the three charges say in a triangle position one would find the force on the test charge as the sum of the two forces from q1 and q2. Otherwise about the shielding of the charge q1 in the case this charge is behind q2 can’t be any doubt.
Now, moving q1 from the triangle position into the shielded position should, thee force between the test charge and q1 has to be a steady function and to be zero in the shielded position.

It is fully ok to be downvoted but in this case I really would be asking to explain what is wrong. The clearness of the above example with the shilded force seems to be without any doubt for me.
