Answer
$n \gt e^{e^{1000}}-3$
Work Step by Step
We have $S_n=a_1+a_2+......+(-1)^{n+1}a_n$
Now, $|S-S_n| \leq |a_{n+1}|$
So, $|a_{n+1} | \lt 0.001 $ or, $|\dfrac{1}{\ln (\ln (n+3))}| \lt 0.001$
or, $\ln (n+3) \gt e^{1000}$
or, $n \gt e^{e^{1000}}-3$
Thus, we need $n \gt e^{e^{1000}}-3$ terms to estimate the sum of the entire series when an error is less than $0.001$.
