Sorry, but how is this answering the question: How does attenuating the level going into the amplifier reduce the load on the preamp?
From
https://www.electronics-tutorials.ws/attenuators/l-pad-attenuator.html
With a potentiometer, R1+R2 is constant, but the output tap varies along the resistor.
R1 + R2 = Rtotal
R1 = Rtotal - R2
R2 = Rtotal - R1
An L-pad uses fixed resistor values.
The output load is the amplifier input Rin and is in parallel with R2, so the effective R2 is always less than the attenuator's R2 value. If the control is set so R1 is not zero (i.e. maximum gain), then R1 is in series with and thus adds to the impedance the preamplifier (Vin) sees. You could also add another fixed resistor in series with R1 so R1 effective becomes R1 + R1series, further raising the impedance, but that will also increase the attenuation.
To use an example, consider a 50k attenuator (potentiometer, pot) so R1 + R2 = 50k-ohms. Assume the amplifier has 50k-ohm input impedance (there will be capacitance as well, dropping the impedance at higher frequencies, but ignore that for now). Now set the pot to the center, attenuating by about 6 dB (1/2 in voltage to the power amp).
That means R1 = R2 = 25k-ohms, but the amp's input puts 50k in parallel with R2 yielding 16.7k-ohms* and overall impedance of (25 + 16.7) = 41.7k-ohms, less than the power amp alone. Note the attenuation is also larger, now about 8 dB, so turning pot to get back to 6 dB will change the values a bit.
Now use a power amp with 10k-ohm input impedance (more like a SS amp) and the same 50k pot, again set to the half-way position. Then R1 = 25k as before, and R2 effective is 25k in parallel with 10k or 7.14k. Now the load the preamplifier sees is 25k + 7.14k = 32.14k-ohms, higher than the power amp alone. But the attenuation is also much greater (-13 dB) so you need to adjust the control to reach 6 dB of attenuation. After some algebra (hopefully correct, no promises), R1 = 8.1k and R2 = 41.9k (remember R1 + R2 = 50k, the overall value of the pot), providing R2 effective = 30.9k // 10k = 8.1k so we get our 6 dB of attenuation. The preamp load is now 16.2k-ohms, still higher than the amp alone.
This is why a simple answer is difficult; the impedances are all interrelated so you must know the amp's input impedance, attenuator's resistance (overall, end-to-end), and setting (attenuation, division range) to calculate the result. I have also ignored the preamp's output impedance as small enough to not significantly change the result.
HTH - Don
* For Ra in parallel with Rb the resulting resistance Rc = Ra * Rb / (Ra + Rb)