Yes, if you power the "5V Pro Mini" with 3.3V. The 16 MHz clock is out of specification, but it has always worked for me. I remove the voltage regulator and power the Pro Mini through the Vcc (or 5V) pin.
The voltages at the inputs/outputs of the microprocessor will depend on the voltage with which the microprocessor will be supplied.
The ATMEGA328PAU microprocessor can be supplied with a voltage between 1.8V minimum and 5.5V maximum. Datasheet-ATMEGA328PAU.pdf
Therefore, we can change the LDO (U2 <=> MIC5205) that equips these cards so that the microprocessor will be powered with 3.3V or 5.2 Volts (or what you choose)
or we can deactivate the LDO (and the led) by removing the soldering of the SJ1 bridge to supply the microprocessor with an external power supply.
I think as long as you remove the SD card while programming, you can use either a 3.3V or a 5V UART adapter so long as the adapter is powering the Pro Mini.
Also, it may depend on which regulator the Pro Mini clone has installed, but it may not be necessary to remove the regulator when powering at its ouput pin (the Vcc pin). With its power indicator LED and regulator removed, a Pro Mini can deep sleep at less than 1uA. If it's not much more than that with the regulator still there, you can just leave it be.
Edit: The Pro Mini clones I've bought don't have that regulator isolation jumper. Apparently the original Sparkfun Pro Mini had it, but I haven't seen it on clones.
Sleep current test:
#include <avr/sleep.h>
#include <avr/wdt.h>
int i;
void setup(){
for (i = 0; i < 20; i++) { // all pins to one rail or the other
pinMode(i,OUTPUT);
digitalWrite(i,LOW);
}
ADCSRA = 0; // disable ADC for power saving
wdt_disable(); // disable WDT for power saving
set_sleep_mode (SLEEP_MODE_PWR_DOWN); // Deep sleep
sleep_enable();
sleep_bod_disable(); // disable brownout detector during sleep
sleep_cpu(); // now go to sleep
}
void loop(){
}