The IfOn Max current is the current that the least sensitive relay should require to turn in - most will turn on at lower currents, but you should supply at least 3 mA to ensure that any relay will be on. The Ablosute Maximum LED current is 50 mA, so it is safe to supply more than 3 mA.
For turn-off, there is IfOff Minimum of 0.4 mA - most units will turn off at a higher current, but the most sensitive may stay on with currents as low as 0.4 mA, so you should ensure that the LED current is below 0.4 mA when you want the thing to be off.
According to the datasheet, the G3MB-202P is a discontinued zero-crossing switched AC SSR.
So, when it is commanded to turn on, it will delay by up to 1/2 of an AC cycle until the next zero crossing. So, at 50/60Hz it can delay by as much as 10msec or 8.33msec
When it is commanded to turn off, it cannot turn off until the current crosses zero. This can be as much as 10msec or 8.33 msec again, but not necessarily in phase with the voltage.
For a resistive load you if you commanded it on for a bit more than 10msec it would be guaranteed to turn on, and commanded it off for a bit more than 10msec it would be guaranteed to turn off. Using say 10.5 msec for each, that's 21msec or about 47Hz. In reality you would get beating between commands and response frequencies with wildly varying amounts of power, so perhaps 1/10 of that frequency or about 5 or 6Hz is more reasonable as a maximum, but even that is a bit high.
If you want to get 'even' and fairly-beat free power (for example for a PWM heater control), normally 0.5Hz is about right (gives you 200 half cycles per period at least).
Nothing you can do by commanding the relay will damage it with a simple load like lights, but you may get undesirable (or perhaps interesting) variations in light if you cycle it too fast. Some will probably appear as smooth pulsations in brightness, which might be quite okay. The life of the lights might be shortened a bit if they're incandescent, but they're only Christmas lights.
Best Answer
Figure 1. From the datasheet we can see that the leakage current is 7 mA.
A 12 W lamp will draw \$ I = \frac P V = \frac {12}{240} = 50 \ \text {mA} \$ (using 240 V to make the maths easy). 7 mA is a significant current for this lamp.
That's the voltage drop across the internal snubber capacitor. You have 180 V across it and 50 V across the lamp.
Be very careful with this. You are working with live circuits. You can't measure resistance on a live circuit and, even if the mains side is dead, you can't measure the resistance of a triac in this way.
All you can normally do is add resistance or another lamp in parallel with the lamp. If you are feeling adventurous you can isolate the power, open up the SSR and snip one of the leads on the snubber. Most will be partially filled with potting compound making this impossible.
Figure 2. See BigClive's Electronically controlled LED lamps glowing when off which explains this well.