The LEDs are designed for a constant current supply which produces whatever voltage is necessary so reducing the current would be effective in dimming the LEDS.
Rohat calculated that the supply output would be a constant 700mA and for 3W per light you need 86mA each. Putting 8 lights in parallel would give 700/8=87.5mA per light which is about right, but that assumes they will share the current equally. The problem is that LEDs have a fairly sharp turn-on so whichever bulb has the lowest forward voltage will take almost all the current and die, then the next lowest would blow and so on. Unfortunately, whichever takes the highest current will heat up most and that will probably drop the voltage futher giving thermal runaway.
What you would need to do is add something to equalise the currents, a series resistor as shown would do but I suspect the value needs to be higher than shown in the circuit, at least the maximum forward voltage variation divided by the current variation you can tolerate. As the LEDs can take up to 350mA, that is fairly easy but you probably want to keep the brightness of the bulbs fairly similar so for example 10% brightness variation implies no more than 8.6mA difference between the bulbs. Higher values of resistor will maintain more even brightness but will increase heat dissipated in the resistors. The advantage is that including 8 resistors is a lot cheaper than separate drivers.
P.S. You probably realise that the supply voltage need only be 35V plus the resistor drop so a 144V-capable supply is much more than you need, it is really intended for driving multiple bulbs in series so that all get exactly the same current and hence brightness.