Driving an LED from a uC pin with the Anode above uC's Vcc rail
Subjecting a I/O pin to voltages above the processor's power level is not a good idea.

Use a transistor:



Added:

The correct solution that doesn't violate any parameters in the datasheet is above.  However, if you are trying to get away with it, then it will most likely work if you make sure the output pin is always driven.

We are assuming a modern processor with a CMOS output.  CMOS transistor can conduct current in both directions when on.  The little bit of current you will get thru the LED when the output is high will be conducted thru the top driver transistor to Vdd.  The value of that current at 1.2 V forward bias will be hard to predict, will vary significantly with temperature, and will vary part to part.  However, it will be "very small" compared to the overall current draw of the processor.

If you switch the output between driven low and high impedance instead of driven low and high, then the protection circuitry for that pin will take the little LED current instead.  For some designs, this is just a diode to Vdd and therefore no worse than having the top FET conduct.  However, other designs have more complicated circuitry which would be worse to dump the current thru.  Guessing what causes the least trouble is one of the problems you run into when ignoring specs.

There are two more ways to be fully in spec:

Add a small bleeder resistor to ground.  Size this to take whatever you determine will be the maximum LED current under all conditions.  The LED may be very dimly lit when it's supposed to be off.

Use a pin that has "5 V tolerant" input, and switch the pin between driven low and floating.