That's not exactly OP's question, but I find it illuminating to look at the explicit formula for $\sum 1/p$. Allow me to work, for simplicity, with $$\sum_{n \le x} \frac{\Lambda(n)}{n\log n}$$ which is closely related to the original sum because it equals $$\sum_{p \le x}\frac{1}{p} + \sum_{p^k \le x, \, k \ge 2} \frac{1}{k p^k }.$$ (The contribution of higher powers can easily be absorbed in the constant $M$ and in the error term.)

There is an explicit formula for ${\sum_{n \le x}}' \Lambda(n)/n^s$ due to Landau (as usual, the $\prime$ in the sum indicates the last term is counted with weight $1/2$ if $x$ is a positive integer). Integrating it with respect to $s$, one obtains $${\sum_{n \le x}}' \frac{\Lambda(n)}{n\log n} = \log \log x +\gamma -\sum_{\rho} \int_{0}^{\infty}\frac{x^{\rho-1-t}}{\rho-1-t}dt$$ where the sum is over all zeros of $\zeta$. So any given zero-free region will give a corresponding error term. The integral should not intimidate anyone, since it can be approximated as $$\int_{0}^{\infty}\frac{x^{\rho-1-t}}{\rho-1-t}dt = \frac{x^{\rho-1}}{(\rho-1)\log x} \left(1+O\left( \frac{1}{\log x}\right)\right).$$ For instance, the Vinogradov--Korobov zero free region will imply $$\sum_{p \le x} \frac{1}{p} = \log \log x + M + O(\exp(-c(\log x)^{3/5}(\log \log x)^{-1/5})).$$

This explicit formula is known to experts, but a bit tricky to find in the literature. Two references:

• Youness Lamzouri, "A bias in Mertens' product formula", Int. J. Number Theory 12 (2016), no. 1, 97–109. In Proposition 2.1 you can find an explicit formula for $\sum_{p \le x} \log(1-1/p)$.
• See Corollary 4.2, with $s=0$, in my arXiv preprint: https://arxiv.org/abs/2211.08973 .

That's not exactly OP's question, but I find it illuminating to look at the explicit formula for $\sum 1/p$. Allow me to work, for simplicity, with $$\sum_{n \le x} \frac{\Lambda(n)}{n\log n}$$ which is closely related to the original sum because it equals $$\sum_{p \le x}\frac{1}{p} + \sum_{p^k \le x, \, k \ge 2} \frac{1}{k p^k }.$$ (The contribution of higher powers can easily be absorbed in the constant $M$ and in the error term.)

There is an explicit formula for ${\sum_{n \le x}}' \Lambda(n)/n^s$ due to Landau (as usual, the $\prime$ in the sum indicates the last term is counted with weight $1/2$ if $x$ is a positive integer). Integrating it with respect to $s$, one obtains $${\sum_{n \le x}}' \frac{\Lambda(n)}{n\log n} = \log \log x +\gamma -\sum_{\rho} \int_{0}^{\infty}\frac{x^{\rho-1-t}}{\rho-1-t}dt$$ where the sum is over all zeros of $\zeta$. So any given zero-free region will give a corresponding error term. The integral can be approximated as $$\int_{0}^{\infty}\frac{x^{\rho-1-t}}{\rho-1-t}dt = \frac{x^{\rho-1}}{(\rho-1)\log x} \left(1+O\left( \frac{1}{\log x}\right)\right).$$ For instance, the Vinogradov--Korobov zero-free region implies $$\sum_{p \le x} \frac{1}{p} = \log \log x + M + O(\exp(-c(\log x)^{3/5}(\log \log x)^{-1/5})).$$ (To be precise, one needs a truncated version of the above formula to deduce this.)

This explicit formula is known to experts, but a bit tricky to find in the literature. Two references, that do exactly as mentioned above (integrate Landau's formula) but with more detail:

• Youness Lamzouri, "A bias in Mertens' product formula", Int. J. Number Theory 12 (2016), no. 1, 97–109. In Proposition 2.1 you can find an explicit formula for $\sum_{p \le x} \log(1-1/p)$.
• Corollary 4.2, with $s=0$, in my arXiv preprint: https://arxiv.org/abs/2211.08973 .