Cramer's conjecture gives the probability that $p_{i+1} - p_i$ is $k$ is like $e^{ - k / \log p_i})/ \log p_i$ so using $$\frac{1}{p_{i+1}^2- p_i^2}= \frac{1}{ (p_{i+1} - p_i) (p_{i+1}+p_i)} \approx \frac{1}{ 2 (p_{i+1}-p_i) p_i}$$

the expected contribution from $p_i$ is

$$ \frac{1}{2 p_i} \sum_{k=1}^{\infty} \frac{e^{ - k / \log p_i}}{ k \log p_i} \approx \frac{1}{2 p_i} \frac{ \log \log p_i}{ \log p_i} $$ (here small $k$, say $k< \log p_i^{1-\epsilon}$, dominate, so we can treat $e^{ - k/\log p_i}$ as the constant 1 supported on $k< \log p_i$) and since the probability that $n$ is prime is $\frac{1}{\log n}$, the expected contribution from $n$ is

$$ \frac{\log \log n}{2 n (\log n)^2 } $$

so a first-order heuristic is $$\int_{e^e}^{\infty} \frac{ \log \log x}{ 2 x (\log x)^2} dx =\int_{e}^{\infty} \frac{ \log y}{ 2 y^2} dy = \frac{ - \log y -1 }{ 2y} ]_{0}^{\infty} = \frac{1}{e}$$

and a heuristic incorporating numerical data is

$$ \sum_{i=1}^{n} \frac{1}{ p_{i+1}^2 - p_i^2} + \int_{p_{n+1} }^{\infty} \frac{ \log \log x}{ 2 x (\log x)^2} dx = \sum_{i=1}^{n} \frac{1}{ p_{i+1}^2 - p_i^2} + \frac{1}{2} \left( \frac{ \log \log p_{n+1} + 1}{ \log p_{n+1} } \right) $$

I don't think the corrections for congruences to small moduli, like forcing all primes to be odd, will affect the leading term here, although they should probably introduce lower-order terms.