All,
Thank you for the explanations! I think I've put the pieces together (for my understanding), and can summarize it as follows:
The potential at the pin is not so much dependent on the external circuitry, but rather is dependent on the voltage drop across the protection diode to Vdd (which is ~zero when powered-down). And THAT voltage drop is dependent on the current going through the diode--which in turn IS controlled through the external circuitry.
In calculating the current from the external circuitry, there is negligible resistance from the diode (ideally), so the only limiting of the current comes from the external resistor. Thus I = V/R.
And the reason that 0.7V above Vdd is bad, is because once the voltage drop across the diode starts going beyond ~0.7V (or 0.3V in case of the PIC16F1783), the current can quickly exceed the diode's limitations and the diode fails.
If I've misstated something, please correct me. Otherwise, thank you for the explanations.