Just 6 digits

Binomial theorem gives us that $(300000-1)^n = (-1)^n + (-1)^{n-1}(300000n) + ...$ where $n = 29999^{2999^{299^{29^{2}}}}$

Subsequent terms will include at least 10 trailing zeros so can be ignored. Clearly n is odd, so $(-1)^n = -1$ ; the last six digits of $300000n$ will be the last digit of $3n$ followed by 5 zeros.

The last digit of $n = 29999^{2999^{299^{29^{2}}}}$ can also be found using the binomial theorem, as $n = (30000-1)^m$ , where $m = 2999^{299^{29^{2}}}$

$n = (-1)^m + (-1)^{m-1}(30000m) + ...$ , and since m is clearly odd, $(-1)^m = -1$ , and so the last digit of n is 9.

Thus the last digit of 3n is 7, and the last six digits of the given number are 700000-1 = 699999