The issue is that when leakage is multiplied by a million transistors in a full-chip digital block, for each digital CMOS pair, one of the transistors has Vds = Vdd, and while the transistor is "off" it still leaks some current, and the lower the threshold voltage, the higher the (Vgs - Vth), so the leakage current goes up. In the current mirror, the circuit is active. There isn't any "leakage" other than the tiny amount going into well or substrate. So the low threshold device serves the purpose of getting you to a higher current with a much lower voltage overhead, which is usually a good thing.