Around these parts, the burden of proof is on the accuser, but you know what, fine.
Most people here know about a resistive voltage divider, and the magnitude of the output voltage is a function of the top and bottom impedances:
So most people know the output voltage Vout wrt. the input voltage Vin: it's just Vout = Vin*(Z2/Z1+Z2).
Of particular note is that if Z1 and Z2 are purely resistive, then there is no frequency component in this equation. If Z1 = Z2, Vout = 0.5 * Vin for
all frequencies...whether or not Vin(t) varies at 1Hz or at a gazillion Hz. In fact, if you need a 0.5 multiplier, there are an infinite number of Z1's and Z2's that will achieve this.
Now what if Z1 and Z2 are not purely resistive, and are in fact purely capacitive? The equation stands, however Z1 becomes Xz1 and Z2 becomes Xz1: in other words, the Z1 and Z2 components now have a frequency component.
Since the exact math bores people: at a high level, we can take the Laplace equivalent circuit, and see that both the numerator and denominator both have a term that is proportional to 1/sC, which means I can simplify by multiplying by the constant term s/s. In short, this means that
the frequency component cancels out of the above equation. This means that the input to output voltage ratio
is constant over frequency, much like the resistive voltage divider. The exact equation reduces down to Vout = Vin*(C1/C1+C2). Notice the difference! Unlike the resistive divider which increases the ratio by increasing the bottom component, for the cap divider the ratio increases by increasing the top component.
So what is the actual difference you might ask? The output impedance of the circuit
does change over frequency, unlike the resistive voltage divider. If the top and bottom caps are 100pF, then at 100Hz it looks (and behaves) like a resistive divider with both resistors equal to 15Mohm. If the frequency increases to 1MHz, then it looks (and behaves) like a resistive divider with both resistors equal to 1.5kOhms. It's like a resistor divider where both resistors scale equally as a function of frequency.
So lets bring this back to microphones: in a microphone, the top cap is the capsule, and the bottom cap is the input capacitance of the tube. In reality, the bottom capacitor is the parallel combination of the grid resistor and the input capacitance. From a frequency response point of view, it doesn't matter: the frequency terms still cancel.
By the capacitive equation above, something interesting emerges: if there were a way to make the bottom capacitance bigger, then the ratio would cause the overall output to reduce. This is the standard way to make a pad: you add a capacitance effectively in parallel with the tube input capacitance, and the
overall output level drops, but the frequency response doesn't change. For example, look at the U67: there is a switch that adds 500pF attached to the output of the capsule, leading down to the backplate, which is sitting at an effective AC ground. This effectively increases the input capacitance of the tube (as seen by the capsule), and provides the pad function. Miller effect does the same thing: the bottom cap increases, which causes the level to drop at all frequencies. I would bet that one could construct a pad with a tetrode, by switching its potential between the cathode and the plate. This would engage and disengage the Miller effect and cause a pad-like effect (all other things being equal, of course).
If this isn't clear enough for anyone to believe, I can also provide a Spice simulation which shows this effect quite clearly.
To summarize: the Miller effect of the tube
does cause a change in level, but with purely capacitive sources, it
doesn't cause a change in frequency response.