You must experiment; first disconnect the resistor and Zobel from the secondary. Feed the primary via a resistor, then increase the resistor until you see significant LF loss. I would start with 200 ohms or so, and a level of about -20 dBu.
There should be a resistor value that creates a 3 dB attenuation respective to midrange; the choice of LF frequency is yours, if you want to design for 20Hz, measure at 20 Hz, if you want to design for 30 or even 40 Hz, do as you please. Let's say with 2400 ohms in series, you have -3dB at 20 Hz, and you want to achieve -1dB, the resistance must be halved, so you must make sure that the source impedance, either direct or via an attenuator, must be less than 200 ohms. In this particular case, some mics would have too high impedance for proper LF response.
This is the basic high-pass effect of an R-L circuit (see attached).
For an attenuator of more than 20 dB, it's the shunt resistor that governs the output impedance. With the example values, that means that the shunt resistor should be smaller than 200 ohms.
That is without loading the secondary. There is no good reason to load the secondary significantly. Using a resistor to damp the xfmr's resonance is brute force. A Zoble is a much more elegant and productive solution.
In a previous post, I explained briefly how to work the Zobel, but the results you presented show that something is wrong in your set-up.
If the secondary is loaded by a resistor, this resistor is reflected at the primary, in parallels, so the effective impedance seen by teh primary is the result of the source Z in parallels with the reflected Z. This may be the reason why the secondary was so heavily loaded originally. It's a dirty trick to make a less-than-decent xfmr look like a half-decent one. Of course, in practice, there is the added attenuation, that often results in less-than-optimum S/N ratio.
