How can i find $Ω(f(n))$ and $O(f(n))$?
It will be easier to analyze $S(n) = \log f(n)$, which is given by the formula
$$ S(n) = \sum_{i=2}^n \log\log i. $$
According to the Euler–Maclaurin formula,
$$
S(n) = \int_2^n \log\log x \, dx + \frac{\log\log 2 + \log\log n}{2} + \sum_{k=1}^s \frac{B_{2k}}{(2k)!} [(\log \log)^{(2k-1)}(n) - (\log\log)^{(2k-1)}(2)] + R,
$$
where the remainder satisfies
$$
|R| = O_s \left(\int_2^n (\log\log)^{(2s+1)}(x) \, dx\right).
$$
Here $(\log\log)^{(r)}$ is the $r$th derivative of $\log\log$ and $B_{2k}$ are Bernoulli numbers.

In order to get big O asymptotics, it suffices for the error to be of order 1. The first few derivatives of $\log\log x$ are
$$
\log \log x, \frac{1}{x\log x}, -\frac{\log x + 1}{x^2\log^2 x}, \frac{2\log^2 x + 3\log x + 2}{x^3\log^3 x}.
$$
The third derivative has constant sign (for $x$ larger than some constant), and so when $s=1$, the growth rate of the remainder term can be bounded by the second derivative evaluated at $n$, which is $o(1)$. So it suffices to choose $s=1$. We obtain
$$
\begin{align*}
S(n) &amp;= \int_2^n \log\log x \, dx + \frac{\log\log 2 + \log\log n}{2} + \frac{\frac{1}{n\log n} - \frac{1}{2\log 2}}{12} + o(1) \\ &amp;=
x\log\log x - Ei(\log x) \bigr|_2^n + \frac{\log \log n}{2} + O(1) \\ &amp;=
n\log\log n - Ei(\log n) + \frac{\log \log n}{2} + O(1),
\end{align*}
$$
where $Ei(x)$ is the exponential integral. The exponential integral doesn't really reduce to any known function, but it is known that $E_i(x) = (e^x/x)(1 + O(1/x))$. Therefore
$$
S(n) = n\log\log n - \frac{n}{\log n} + O\left(\frac{n}{\log^2 n}\right).
$$
This gives the value of $f(n)$ up to a factor of $e^{O(n/\log^2 n)}$. In the same way, we can get an explicit expression with error $e^{O(n/\log^k n)}$ for any constant $k$, using the divergent series for $Ei(x)$. If we want an expression with $O(1)$ multiplicative error, then the best we can get is
$$
f(n) = \Theta(e^{n\log\log n - Ei(\log n)}\sqrt{\log n}).
$$
With a bit of effort we can also find the hidden constant.