Can you count time as a parameter?
No. The configuration is what potentially changes over time. So the equations of motion are a function from time, into the set of configurations. Time is the domain and the set of configurations is the codomain and the equations of motion is the function from the domain to the codomain.
Can't you always increase the number of parameters (even if it has no effect) and still determine the configuration of the system?
They would not be independent.
In the case of a pendulum, most texts say it has only one degree of freedom,
That's the planar pendulum, confined to rotate in a plane, like a grandfather clock.
In the case of projectile motion, the projectile has 3 degrees of freedom, right?
Yes. At each point in time you have to specify three coordinates to specify the configuration at that time. More if it is extended and can have orientation, even more if it is not rigid.
Keep in mind that a degree of freedom is about the space of possible configurations. It isn't about any one particular equation of motion.
Can someone give me a rigorous definition of the degrees of freedom and explain how this definition addresses the questions above?
The only place it seems you stumbled is about independence. You should be able to freely adjust any of the coordinates in the degree of freedom within some little bit and have a different configuration.
But this is also a false generality. For $N$ particles the degrees of freedom is $3N$ and sometimes you can pretend there are fewer by pretending that some constraint is exact when it is not actually exact. For instance a real pendulum can and does elongate a little bit and the place it pivots can wiggle a little bit and so forth. The one degree of freedom is really about ignoring the other degrees of freedom.
So just have $3N$ and then start eliminating ones you don't care about whose dynamics hardly change in an important way. And just don't over eliminate, you should retain enough to describe your system. In the case of the pendulum when you know the end point and you assume the rigidity and one part fixed, then you know the whole thing.
What can you really gain by pretending to have more generality than is really there?
We should start by acknowledging that "degree of freedom" has at least two meaning in physics and engineering terms. One—used in mechanical design—is essentially equivalent to a mechanical generalized coordinate for the system. Another—the more common meaning in thermal physics—is the dimensionality of the system's phase space.
Neither of these thing is quite what you should be counting for the purposes of the equipartition theorem.
The thing that the equipartition theorem counts is contributions to the Hamiltonian that are quadratic in either a generalized coordinate or a generalized momentum.
A 1D spring has the freedom to move in only one direction (i.e. one mechanical degree of freedom using the engineering definition), but has a two dimensional phase space $(x,p_x)$ (i.e. two degrees of freedom in a common but to my ear sloppy usage), and the Hamiltonian is quadratic in both parameters
$$ H = \frac{1}{2}k x^2 + \frac{p_x^2}{2m} \,,$$
so it has two contributions for the purposes of equipartition.
In a model solid like the one you exhibit each atom (in a $D$ dimensional space) can be naively associated with $D$ mechanical degrees of freedom, $2D$ parameters in phase space, and $2D$ quadratic modes in the Hamiltonian.
As a side note it is not necessarily true that each mechanical degree of freedom results in two contributions to the equipartition theorem. The most handy counterexample being the ideal gas.
Best Answer
Assuming that the $N$ argon atoms are being treated as a classical system, then there are 3 degrees of freedom per atom. That assumes that we are neglecting electronic degrees of freedom, which is OK since one needs a fairly high temperature to thermally excite the electrons in argon.
Now if there are $N$ atoms, then there are $3N$ degrees of freedom. No if ands or buts. However, when doing computer simulations it is very common to make the center of mass of the system of atoms stationary. The center of mass can always be separated out from the other coordinates of the system and so can its coordinates and momentum. In this case there are $3N-3$ remaining degrees of freedom.
The reason why this is often done in computer simulations is that what is interesting are the motions inside the gas, and not the bulk motions of the gas as a whole. More importantly, the center of mass will remain fixed and the momentum of the center of mass will be zero. To see this understand that there are no forces acting upon the center of mass.