Simple fix: reposition C2. Connect the minus side to the minus of C1. Connect the plus to pin 13 of the IC ( where the minus used to be. ) This will pull the gate low when the circuit is first powered up. On further thought; I think the connection of R2 will also need to be changed. Cut the trace that connects to the FET and add a jumper so it ,R2, connects to the pins 12-11 on the IC. The FET will turn on just enough to stop charging C1 so first blink might be dim. This seems to be the most doable change.
Parduz reports that this didn't do the trick so new configuration. Move the plus side of C2 to the junction of R2 and R3. I can't read the value of R3; if it is 75K it can stay, but it is 7.5K it will need to be replaced with a resistor between 47K to 220K; whatever you have or can obtain. Then put a new capacitor in the spot that used to have C2; value will depend on the new R3, if 100K then .33uF to .47uF, .1 to .22 uF for 220K, etc. This will hold the FET off for 30 to 50 mSec, allowing C1 to charge up enough to fully turn on the FET. This won't address the complaint of too much variation in timing with voltage change, Unfortunately I am not in a position to completely re-engineer this circuit. This is the most I can think of for now.
Further thoughts: I have been assuming that C1 discharges when the blinker is off, but now realize that is not likely. The oscillator will always stop in the off state and to get it to reset, both C1 and C2 must go to zero volts. To do this a resistor should be connected across C1; 47K would would discharge it in about 15 seconds while causing about 10% ripple when on. This is likely to make the timing variation with voltage worse, so it will be a trade off.