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