I would like to know if there is a result for the product $$f(x)=\prod_{p\leq x}\log p,\quad \text{where $p$ is prime}.$$
A simple upper bound is $f(x)<(\log x)^{\pi(x)}$, where $\pi(\cdot)$ is the prime counting function. But I'm hoping to find a closer approximate to the function.
The Chebyshev's theta function $\displaystyle\theta(x)=\sum_{p\leq x}\log p$ is the closest function I can find.
$(\log x)^{\pi(x)}$ is actually pretty close to the truth, because $\log p \approx \log x$ for the vast majority of the primes up to $x$.
The way you'd approach all the information you want about the quantity is: \begin{align*} \log f(x) = \sum_{p\le x} \log\log p &= \int_{2-}^x (\log\log t) \,d\pi(t) \\ &= \pi(x) \log\log x - \int_2^x \frac{\pi(t)}{t\log t}dt. \end{align*} (The penultimate expression is a Riemann-Stieltjes integral, but you can ignore it and even prove by hand that the second and fourth expressions are equal.) From this you can see for example that $\log f(x) = \pi(x) \log\log x + O(x/\log^2x)$. Exponentiating you get a lower bound like $f(x) > (\log x)^{\pi(x) - Cx(\log\log x)/\log^2x}$ for some constant $C>0$.