Reading the question and the comments, there may be a conceptual misunderstanding : the attenuator WILL attenuate any noise presented on its input (even from just a 50 ohm source impedance), to the same extent it attenuates the signal.
However it also generates noise of its own, which may be represented as the noise from a perfect resistor equal to its own output impedance, and this is added at the output to the (attenuated) input signal and noise. So if input and output Z are both 50 ohms, the net result is attenuated signal + marginally increased noise (i.e. NF = attenuation).
But if its output impedance is lower, the added noise is also lower, thus improving the noise voltage as Andy states.
So represent the attenuator as a perfect attenuator (attenuating noise) in series with a Johnson noise voltage source equal to the output impedance. The rest is just applying the formulae.
EDIT: re: updated question.
(1) There is nothing special about 290K except that it's a realistic temperature for the operation of a passive circuit. The reason they chose it is that the article quotes a noise floor ( -174dBm/Hz) which is correct for a specific temperature : yes, 290k.
(2) While any resistance in the attenuator will contribute noise, I realise that it is not a satisfactory explanation as to why you get the same noise out of an attenuator, because (as Andy says) you could make a capacitive attenuator which is not a Johnson noise generator. So we have to look a little deeper, and remember these noise sources are the statistics of the individual electrons that make up the current.
So, let's say we build a (50 ohm in, 50 ohm out) attenuator, and attempt to cheat Johnson by using a capacitive divider. That implies a node within the attenuator which conducts some of the input current to ground. At this node, we have two current paths; a fraction of the current flows to output, the rest to ground. What determines which path an individual electron will take? Essentially, chance. Collectively? Statistics. So this is a noise source.
Or let's just add series capacitance to provide enough attenuation : we thereby avoid dividing the current flow and eliminate the noise source, right? At the cost of reducing the signal current; our statistics now operate with a smaller sample size and consequently greater variance : more noise.
These results are the best you can do, there is no way round them.
Yes, the divider definitely will affect the gain calculation, because its equivalent source resistance (Thévenin resistance) of 6.23 kΩ is in series with one of your 10 kΩ gain-setting resistors. The overall gain will be -0.2325, rather than the -0.375 that you're looking for.
If you want an overall gain of -3/8, it would be simpler to forget about the voltage divider and just set up your opamp with 20 kΩ input and 7.5 kΩ feedback resistors.
Also, be sure to use an opamp whose common-mode input range includes ground.
Best Answer
I assume that 6K is intended to represent the input impedance of the ADC, so it doesn't contribute noise.
The R5/R6 combination contributes 2.0nV/sqrt(Hz) at the ADC input. The output of the op-amp has noise of about 18nV/sqrt(Hz) (mostly due to Johnson noise in the resistors), so about 1.9nV/sqrt(Hz) at the divider output, or a total of 2.8nV/sqrt(Hz) at the ADC input. In other words the two contributions are similar (divider and amplifier).
I encourage you to do a full noise analysis and look at the various contributions individually. Since they add in quadrature, the magnitudes of the squared values are important. This Analog Devices paper gives you the information you need.
I may have made an error, but when I do the analysis with an attenuator made with 10K/546.5 ohm resistors rather than 10K/5K and no divider (gain is 0.0546 in both cases), I get a noise for your circuit of 51nV/sqrt(Hz) referred to the input of the diffamp vs. 119nV/sqrt(Hz) for the no-divider circuit, corresponding to 2.8nV/sqrt(Hz) vs. 6.49nV/sqrt(Hz) at the ADC input.
All this stuff refers to the region above the voltage and current noise corners, of course, which should be valid for most of the audio spectrum (but it never seems to be for the stuff I do).