If you're using a series regulator (of which the dropping resistor would be a crude version), your power dissipation would be the line voltage times the average current, assuming a unity power factor--which the dropping resistor provides. So if we assume a 50% duty cycle 20mA setup, the power dissipated is about 1.2W--certainly enough to get rather warm. If it's 20mA average, then of course double the power consumption.
Basically, the LED would be dissipating whatever voltage it's running at, and the resistor the rest to reach the line voltage. The vast majority of the power dissipated is by the resistor--90% if the LED is running at 12V.
You still have to convert AC to DC for the LED somehow, and regulate the voltage it sees when it's off. (Off or reverse biased implies virtually no current consumed, which implies its instantaneous impedance is high, which means the voltage divider with the dropping resistor makes the LED subject to nearly the full line voltage. I predict it would very soon be permanently off.)
I'm reminded of the old cheaper tube radios and TVs that used a mains dropping resistor to control current through the heaters. There are good reasons why a filament transformer was preferable and safer, some of which would also apply here. Those dropping resistors often didn't hold up so well over long time use--maybe modern ceramic resistors are more robust.
