When compared against the available battery voltage, the voltage drop is rather significant.

>> (1.6/10.4)*100

ans =

15.3846


Some 15%

However, as the power in the glow plugs can be written as V^2 / R

>> 100*(1-((10.4-1.6)^2)/(10.4^2))

ans =

28.4024

Compared with a system with no voltage drop, there's 28% less power available to heat the prechamber. (subject to the assumption that R doesn't change)

Obviously, you'll never get a voltage drop of zero, 1.6 volts is excessive. 0.3 volts would be good.