What is the entropy of mixing a gas into the whole atmosphere?
Performing an actual calculation could be quite complicated, even if we use the ideal gas approximation, because the atmosphere has enormous variations in both temperature and density.

To simplify the problem, let's think of the atmosphere as an ideal gas of uniform temperature and density.   Then we need to determine how the entropy of the argon released into the atmosphere increases as a result of this release.

Since we're doing an ideal gas approximation, the argon doesn't "know" about the other atmospheric gases – the only thing that affects its entropy is the increase in volume.  But, when doing an entropy calculation, we do need to account for the fact that the atmosphere already contains 0.93% Ar.   Effectively, the presence of the existing argon removes a fraction of the configurational possibilities, since if we replace argon with argon, we don't change the entropy.   Hence the entropy increase is calculated from the effective configurational volume that remains for the argon relased from the box. 

Specifically, for the purpose of calculating entropy, the effective final volume for the argon released from the box is:

$$V_\mathrm f = V_\text {atmosphere} - V_\text{Ar in atmosphere}+ V_\text {Ar in box}\approx V_\text {atmosphere} - V_\text {Ar in atmosphere} = (1- x_\text {Ar})V_\text {atmosphere},$$
where $x_\text {Ar}$ is the mole fraction of argon in the atmosphere, which is 0.93% = 0.0093.

Proceeding, we start with the following general expression for a pure substance (again, the ideal gas approximation allows us to ignore the non-argon molecules):   

$$\left(\frac{\partial S}{\partial V}\right)_{\!T} =\frac{\alpha}{\kappa},$$
where $\alpha$ is the coefficient of thermal expansion, and $\kappa$ is the isothermal compressibility.

For an ideal gas, $\alpha = \frac{1}{T}$ and $\kappa = \frac{1}{p}$, hence:

$$\left(\frac{\partial S}{\partial V}\right)_{\!T} =\frac{p}{T}$$

If we additionaly assume the expansion is isothermal (i.e., T of the box is the same as that of the atmosphere--which we've already assumed is of uniform T) we obtain, for an ideal gas:  

$$\mathrm dS = \frac{p}{T}\,\mathrm dV = \frac{n R}{V}\,\mathrm dV \Rightarrow \Delta S = n R \int\limits_{V_\mathrm i}^{V_\mathrm f}\frac{\mathrm dV}{V} = nR \ln\frac{V_\mathrm f}{V_\mathrm i}=nR \ln\frac{0.9907 \,V_\text {atmosphere}}{V_\mathrm i}$$

where $V_\mathrm i$ is the initial volume available to the argon, and $V_\mathrm f$ is the final volume available to the argon. 

As you can see, releasing a mole of argon into the atmosphere will result in a significant entropy increase, once the argon is fully dispersed.  

Note that, in the case where the atmosphere is 100% Ar:
$$V_f = V_\text {atmosphere} - V_\text{Ar in atmosphere}+ V_\text {Ar in box}=V_{\text{Ar in box}} \Rightarrow \frac{V_f}{V_i}=1\Rightarrow \Delta S = 0, \text{ as expected.}$$ 

We can also calculate the corresponding free energy change using:

$$\Delta G = \Delta H - T\,\Delta S$$

With the ideal gas approximation, there is no enthalpy of mixing, hence:

$$\Delta G = -T\,\Delta S = -nRT \ln\frac{V_\mathrm f}{V_\mathrm i}.$$

If we assume STP ($273.15\ \mathrm K$, $1\ \mathrm{bar}$), and take the volume of the atmosphere to be the area of the earth's surface times, say, 50 kilometers*, then we have:

$$\Delta S = (1 \ \mathrm{mol})\ R\ \ln\frac{2.53 \times 10^{22}\ \mathrm L}{22.7\ \mathrm L} = 0.403\ \frac{\mathrm{kJ}}{\mathrm K}$$

$$\Delta G = - (1 \ \mathrm{mol})\ R\ (273.15\ \mathrm K) \ln\frac{2.53 \times 10^{22}\ \mathrm L}{22.7\ \mathrm L} = -110\ \mathrm{kJ}$$

*This is all arbitrary, since we're unrealistically assuming constant pressure and temperature; but I needed to pick a number for the calculation, so I chose the top of the stratosphere ($50\ \mathrm{km}$).  $94\ \%$ of the atmosphere's gases are found below this altitude, so about $94\ \%$ of the argon would end up there as well.  

Then again, because of the logarithm, the choice of atmospheric volume doesn't significantly affect the calculation, as you can see if you compare my answer with Buck Thorn's. Indeed, even if we allow the argon to expand to fill the observable universe ($4 \times 10^{83} L$), this only increases $\Delta S$ and $\Delta G$ by a factor of four! [$\frac{83}{22}\approx 4$.]

Calculation done in Mathematica (in Mathematica, ln is represented by "Log"):