The accepted answer you linked to doesn't quite tell you what you need to know... it just shows you how to work out the temperature of the atmosphere (given some simplifying assumptions), given the average surface temperature of the planet... this doesn't tell you how to work out the average surface temperature of the planet given some simple atmospheric model and the equilibrium temperature of the planet (eg. if the atmosphere wasn't there).
Happily, the book they reference does tell you this information. Their link is broken, but a brief search show you its new home: Introduction to Atmospheric Chemistry, Chapter 7.
Lets start with the simple no-atmosphere model of planetary equilibrium temperature: $$T_{eq} = \left[ {F_s(1-A)} \over {4\sigma} \right]^{1 \over 4}$$
$F_s$ is the solar flux at the orbital radius of your planet, in watts per square meter. For Earth, this is ~1361W/m2. You should be able to work out how to compute this yourself easily enough, so I won't go into that here. This model assumes that all the heating of the planet comes from the energy of the sun... geothermal effects are ignored.
$A$ is the planetary albedo, a measure of how much of the incident solar flux is reflected back into space. This is a slightly awkward thing to measure, especially on a planet with varying colors like Earth, but the book uses a value of 0.28. This is a little lower than the Bond albedo of the Earth (~0.304) for reasons I'm too lazy to find out.
$\sigma$ is the Stefan-Boltzmann constant. It and the one-fourth power come from the Stefan-Boltzmann law, which describes the energy emitted by a black-body radiator. This is used to model the amount of energy being radiated away by the Earth, with the planetary equilibrium temperature being that at which the energy being radiated away by the Earth exactly balances the energy being received from the Sun.
The version of the equation you probably want is this one, 7.16, from page 127 of the book (or page 15 of the chapter's PDF):
$$T_o = \left[ {F_s(1-A)} \over {4\sigma(1-{\mathcal{f} \over 2})} \right]^{1 \over 4}$$
The extra parameter $\mathcal{f}$ models the fraction of the energy being radiated by the Earth that is captured by the atmosphere. This obviously warms the atmosphere, which will then in turn re-radiate its heat following the same simple black-body radiator model we've used before. Half of the heat being re-radiated goes back down to the Earth, the other half shoots off into space, which probably explains the 1/2 bit.
Earth's atmosphere can be modelled with $\mathcal{f}$ as being ~0.77. Throwing that value into the equation gives you an observed surface temperature $T_o$ of ~288K, which is close enough for your needs, I'm sure.
It isn't at all easy to determine the value of $\mathcal{f}$ from the composition of the atmosphere, but you can work it out by looking at the observed surface temperatures of bodies in the Solar System (such as Venus or Mars) and using their observed albedos and the solar flux which you can compute yourself. The you can apply your $\mathcal{f}$-values to your fictional worlds. Job done.
The other answer (the non-accepted one) gives a simpler looking model, the Milne-Eddington approximation, which hides most of the details behind $T_{eq}$ but otherwise follows a pretty similar pattern (which shouldn't come as a surprise, because it is trying to model the same thing in the same way):
$$T_o = T_{eq} \left(1 + \frac{3}{4} \tau\right)^{\frac{1}{4}}$$
Where $T_{eq}$ is the planetary equilibrium temperature described above.
This replaces the $\mathcal{f}$-parameter with a related $\tau$-parameter, which is useful because the reference given, Planet temperatures with surface cooling parameterized, gives you a model for $\tau$ using atmospheric $\ce{CO2}$ and $\ce{H2O}$:
$$ \tau = 0.025P_{\ce{CO2}}^.53 + 0.277P_{\ce{H2O}}^.3$$
where $P_{\ce{CO2}}$ and $P_{\ce{H2O}}$ are atmospheric carbon dioxide and water vapor partial pressures in pascals.
On Earth, these values are 33.6Pa and 392Pa respectively, giving a $\tau$ of 1.82. Unfortunately the model is incomplete, because this gives a surface temperature of >315K, which is a little warm. The paper goes on to talk about other details of optical thicknesses and convective cooling and so on, but it doesn't close with a nice simpler formula that would fit your needs. You'd need to read and digest the paper yourself to learn about the conclusions they came to.
