Let$\ p_n$ be the$\ n$-th prime. Is$\ \lim_{n\to\infty} \log \log n \prod_{i=1}^{\lfloor \log n \rfloor} \frac{p_i-1}{p_i}>0$? I'm less than a novice in analysis, I don't even know how to approach this. Thanks in advance.
The limit is $e^{-\gamma}$ where $\gamma$ is the Euler–Mascheroni constant.
This is a direct consequence of


  Mertens' $3^{rd}$ theorem
  $$\lim_{x\to\infty} \log x \prod_{\substack{p\,\le\,x\\p \text{ prime}} }\left(1-\frac{1}{p}\right) = e^{-\gamma}$$


By Prime number theorem, we know $p_n \sim n\log n$, so 
$$\lim_{n\to\infty}\frac{\log n}{\log p_n} = 1$$

If we substitute $x$ by $p_n$ in Mertens' theorem and multiply its LHS by $\displaystyle\;\frac{\log n}{\log p_n}$, we get

$$
\lim_{n\to\infty} \log n \prod_{i=1}^n \left(1 - \frac{1}{p_i}\right)
= \lim_{n\to\infty} \frac{\log n}{\log p_n} \left[ \log p_n \prod_{i=1}^n \left(1 - \frac{1}{p_i}\right)\right]\\ 
= \lim_{n\to\infty} \log p_n \prod_{\substack{p\,\le\,p_n\\p\text{ prime}}}\left(1 - \frac{1}{p}\right) =
 e^{-\gamma}$$

For a proof of Mertens' theorem, see 


G.H Hardy's and E.M. Wright,
Chapter XXII - The Series of Primes (3),
An Introduction to the Theory of Numbers


It is theorem $429$ there.
No, because as $n \to \infty$, we have that $x_n=\prod_{i=1}^{\log n} \frac{p_i-1}{p_i}=\prod_{i=1}^{k} \frac{p_i-1}{p_i}$ diverges*, where $k=\log n \to \infty$.  Hence, we have:

$$\displaystyle \lim_{n \to \infty}\log \log nx_n=\log \log \infty=\infty$$

*(by your other question Let$\ p_n$ be the$\ n$-th prime. Can you give me a proof for$\ \prod_{i=1}^\infty \frac{p_i-1}{p_i}=P\approx \frac{1}{11.0453}$?)