The problem is that when you request a 1 ms tick rate, the kernel may not be able to actually give it to you because of the frequency of the input clock to the timer hardware. In such cases, it chooses the closest number that's faster than what you requested. In terms of IBM PC hardware, requesting a 1 ms tick rate actually gets you 999,847 nanoseconds between each tick. With the requested delay, that gives us the following:
Since the kernel expires timers only at a clock interrupt, the timer expires after ceil(2.000153) ticks, so each delay(1) call actually waits:
999,847 ns * 3 = 2,999,541 ns
Multiply that by a 1000 for the loop count, and you get a total loop time of 2.999541 seconds.