When taking the base-$2$ logarithm, you get a sum that converges faster than the series for $e$.
The first iterates are
$$0.5 \\
0.666666666667 \\
0.7 \\
0.704761904762 \\
0.705194805195 \\
0.705228105228 \\
0.705230064054 \\
0.70523016715 \\
0.705230171632 \\
0.705230171787 \\
0.705230171792$$
When you truncate the sum, the sum of the tail will not exceed twice the first omitted term*. Then
$$S_n>S_\infty-\frac2{p_1\cdot p_2\cdots p_{n}}.$$
From the asymptotics of the primorial,
$$S_\infty-S_n=\frac{e}{n^{n(1+o(1))}}.$$
*A better approximation of this factor $2$ is $1+S_\infty$, or better (and much closer to $1$), $1+\dfrac2{p_{n+1}}$.