For arbitrary $m \in \mathbb{N},$ $$\sum_{n=1}^{m}\ \sum_{d | \#_n}\mu(d)=\sum_{n=1}^{m}\big | \sum_{d | \#_n}\mu(d)\ \big |\ = \ 0,$$ $$\prod_{n=1}^{m}\ \prod_{d | \#_n}d^{\mu(d)}=\prod_{n=1}^{m}\big | \prod_{d | \#_n}d^{\mu(d)} \big |\ = \ 1,$$ where $\#_n:=\prod_{i=1}^n p_i$.
Are these partial sums and partial products absolutely convergent?
If you're asking about the infinite sum $$ \sum_{n=1}^\infty \bigg( \sum_{d\mid\#_n} \mu(d) \bigg) $$ and the infinite product $$ \prod_{n=1}^\infty \bigg( \prod_{d\mid\#_n} d^{\mu(d)} \bigg), $$ then yes, they are absolutely convergent and their values are $0$ and $1$, respectively, by your computations.
I put parentheses around the inner terms to emphasize that this answer is true only when the expressions are interpreted in this (natural) way. For example, the sum $$ \sum_{n=1}^\infty \sum_{d\mid\#_n} |\mu(d)| $$ certainly does not converge; and the double product $$ \prod_{n=1}^\infty \prod_{d\mid\#_n} d^{\mu(d)} $$ cannot have its terms arbitrarily rearranged and still converge.