Slight variation of Mertens' third theorem. Do we have an estimate of $\sum_{d \mid n\#}(-1)^{\omega(d)}\dfrac{\sigma_0(d)}{d}$? Define $f(n)$ to be:
$$
\sum_{d \mid n\#}(-1)^{\omega(d)}\dfrac{\sigma_0(d)}{d}
$$
But $\sigma_0(d) = 2^{\omega(d)}$ for any $d \mid n\#$ a primorial, so:
$$
f(n) = \prod_{p \text{ prime} \\ p \leq n} \dfrac{1}{\left(1 - \dfrac{2}{p}\right)}
$$
Mertens' third theorem has as $1$ instead of a $2$ there.
So what would be an asymptotic estimate of $f(n)$?
 A: We have
\begin{align*}
\prod\limits_{2<p \le n} {\left( {1 - \frac{2}{p}} \right)^{ - 1} } & = \exp \left( { - \sum\limits_{2<p \le n} {\log \left( {1 - \frac{2}{p}} \right)} } \right) \\ & = \exp \left( {\sum\limits_{2<p \le n} {\frac{2}{p}}  - \sum\limits_{2<p \le n} {\left[ {\frac{2}{p} + \log \left( {1 - \frac{2}{p}} \right)} \right]} } \right).
\end{align*}
Now by Mertens' second theorem
$$
\sum\limits_{2<p \le n} {\frac{2}{p}}  = 2\log \log n + 2M -1+ \mathcal{O}\!\left( {\frac{1}{{\log n}}} \right),
$$
where
$$
M = \gamma  + \sum\limits_p {\left[ {\frac{1}{p} + \log \left( {1 - \frac{1}{p}} \right)} \right]} 
$$
is the Meissel–Mertens constant. Also
\begin{align*}
\sum\limits_{2<p \le n} {\left[ {\frac{2}{p} + \log \left( {1 - \frac{2}{p}} \right)} \right]} & = \sum\limits_{p>2} {\left[ {\frac{2}{p} + \log \left( {1 - \frac{2}{p}} \right)} \right]}  - \sum\limits_{p > n} {\left[ {\frac{2}{p} + \log \left( {1 - \frac{2}{p}} \right)} \right]} 
\\ & = \sum\limits_{p>2} {\left[ {\frac{2}{p} + \log \left( {1 - \frac{2}{p}} \right)} \right]}  - \sum\limits_{p > n} {\mathcal{O}\!\left( {\frac{1}{{p^2 }}} \right)} \\ &
 = \sum\limits_{p>2} {\left[ {\frac{2}{p} + \log \left( {1 - \frac{2}{p}} \right)} \right]}  + \mathcal{O}\!\left( {\frac{1}{{n\log n}}} \right).
\end{align*}
Taking exponentials and simplifying
$$
\prod\limits_{2<p \le n} {\left( {1 - \frac{2}{p}} \right)^{ - 1} }  = C(\log n)^2 +\mathcal{O}(\log n),
$$
where
$$
C = \frac{{e^{2\gamma } }}{4}\prod\limits_{p > 2} {\left( {1 + \frac{1}{{p(p - 2)}}} \right)} = \frac{{e^{2\gamma } }}{4\Pi_2}=1.201303559967362\ldots,
$$
with $\Pi_2$ being the twin prime constant.
A: Asymptotically,
$$g(n) = \prod_{2 < p < n}^{p \text{ prime}} 1-\frac2p \sim \frac{4C_2e^{-2\gamma}}{(\log n)^2} \approx \frac{0.83244}{(\log n)^2}$$
And presumably your expression is simply the reciprocal of that, i.e.,
$$f(n) \sim \frac{e^{2 \gamma}(\log n)^2}{4C_2} \approx 1.2013(\log n)^2$$
where $C_2$ is the twin prime constant:
$$C_2 = \prod_{p>2}^{p \text{ prime}} \frac{p(p-2)}{(p-1)^2} \approx 0.66016$$
Note: or just look at the other answer, which does a better job and shows the full derivation.
