Zener noise is variable depending on the device fabrication and parameters, so you can't in general get a fixed noise level from a zener in the same way that you can just order a 5.6V Zener and know that it'll produce that certain breakdown voltage with whatever tolerance is applicable. Of course you can buy characterized and specified purpose built noise diodes which will be guaranteed to produce a certain characteristic of noise under a given set of operating conditions but these are rare or expensive or not necessarily applicable depending on how you want your noise shaped.
The other commenter is correct that you can use a noise source and a variable or selected attenuator to produce a given maximum noise level from a source or some attenuated value from that source, though the source noise density and bandwidth itself might vary depending on age, voltage, current, temperature, load impedance, et. al. so you'll need to measure and select a configuration to produce the appropriate level output for a given source and operating condition.
A common way to adjust the level of a signal would be to use an AGC amplifier such that you apply a variable gain or loss to an input signal such that the power or peak or envelope of the output signal is at some predefined level, and the "automatic" part of the "gain control" will act as a control system to keep the level within your desired setpoint. There are many AGC amplifiers for RF / IF applications that would be applicable -- check Analog Devices, Linear Technology, TI, AVAGO and similar vendors for part options.
Many kinds of resistors have theoretically predictable thermal noise characteristics depending on the measurement bandwidth, resistor temperature, and resistance value, though the noise level is generally much lower than a good noise diode, and often controlled relatively high temperatures are required to generate a lot of noise (relative to other options).
https://en.wikipedia.org/wiki/Johnson%E2%80%93Nyquist_noise
You could use a peseudo random digital generator made by a CPLD, MCU, or other logic device implementing a LFSR or other pseudo random sequence generator and feed that output through whatever kind of DAC, filter, and buffer you require to get a fixed level noise output -- the DAC would generate a predictable output level and the filter / buffer would have known spectral shape and gain/loss.
I suggest a properly biased zener diode with appropriate shielding and construction, in an oven if necessary, followed by an AGC amplifier and filter.
If I recall correctly you could probably find some application notes about such a setup from places like LINEAR TECHNOLOGY, AVAGO, the old AGILENT / HP diodes/discretes ANs, BSTJ, probably M/A-COM, maybe FAIRCHILD or MICROCHIP or ON-SEMI / old Motorola. Maybe the old National Semiconductor linear applications too.
If you used a small FPGA plus DDS or Sigma Delta DAC you could probably generate a selectable level / spectrally colored noise (if that matters to you) relatively easily.
Edit -- more information --
http://cds.linear.com/docs/en/application-note/an61fa.pdf
(see pages 24-26, and appendix b of an61)
http://www.linear.com/docs/4262
http://www.maximintegrated.com/app-notes/index.mvp/id/3469
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.
Best Answer
The signal-to-noise ratio needed for \$\mathrm{32\ kbit/s}\$ with a bandwith of \$\mathrm{100\ kHz}\$ is calculated in the following way:
\$\underbrace{C=B\cdot \log_2{\left(1+\dfrac SN\right)}}_{\textrm{Shannon's equation}}\longrightarrow \dfrac CB=\log_2{\left(1+\dfrac SN\right)}\longrightarrow 2^\frac{C}{B}=1+\frac SN\longrightarrow \frac SN=2^\frac CB - 1\$
If we substitute the values in the equation we get:
\$\dfrac SN=2^\frac{32\cdot 10^{3}\ \textrm{bits/s}}{10^5\ \textrm{Hz}} - 1=2^{0.32} - 1=0.2483\longrightarrow 10\cdot\log_{10}{(0.2483)}=-6.05\textrm{ dB}\$.
In Shannon's equation:
Sources:
https://en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem
https://www.st-andrews.ac.uk/~www_pa/Scots_Guide/iandm/part8/page1.html