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.
Where does 8.28e-19 come from?
Thermal noise of a 50 ohm resistor in a 1 Hz bandwidth at 27 degC is 9.1e-10 volts
To convert this to an equivalent power it needs squaring and this produces a number of 8.28e-19.
Thermal noise calculator.
The formula also reduces to 20 log\$_{10}(\frac{V_{NOISE}}{9.1e-10}\$) i.e. it compares actual RMS noise against 1Hz-limited voltage noise from a 50 ohm resistor.
Best Answer
Searching for similar devices only yields NSVJ3910 (again from ONSemi with a similarly braindead datasheet) and no expected 2N3xxx equivalent (the 2N3819 was a famous example, but I see no 2N3910), leaving 4 options:
(1) Guess. As it's an RF app, I'd guess 50 ohms, but confirm by option (2) or (3).
(2) Measure. Build an amplifier using one, and feed it from (a) a dead short and (b) different resistances. The resistance that increases the output by 3dB is the effective noise resistance of the FET. From that, computing the noise figure for any given source impedance is straightforward.
(3) Ask the supplier's FAE for clarification. On a £0.10 part I'll be highly impressed if you get a competent answer.
(4) Choose a different part, with a better datasheet. Motorola datasheets often struck me as weirdly incomplete next to National Semiconductor or Philips, and On Semi seem to be continuing the tradition... Look at 2N3819 or J309 (or SOT23 equivalents) or 2N5459 if low noise is a priority. Datasheets that show schematics of the test circuit used are usually a good sign...