Answer to question 2: obviously $f(n)\geq 3$ for all $n$, but I will show that $f(n)=3$ for infinitely many $n$, so we cannot give any other lower bound for the growth of $f$. To see this, notice that if $p_2...p_{n+1}\not \equiv \pm 1 \pmod 8$ (where $p_i$ is the $i$th prime), then the highest power of $2$ in $O_n^2-1$ is $2^3$. On the other hand, if $p_2...p_{n+1}\equiv\pm 1 \pmod 8$ and $n$ is chosen so that $p_{n+2}\equiv \pm 3 \pmod 8$ (by Dirichlet's theorem, there are infinitely many such $n$), we get $p_2...p_{n+2}\not \equiv \pm 1 \pmod 8$, so in any case there are infinitely many $n$ for which $f(n)=3$. We can even say something about the density of such $n$. By denoting $A=\{n:f(n)>3\}$, $B=\{n:f(n+1)=3\}$ and $C=\{n:p_{n+2}\equiv \pm 3 \pmod 8\}$, the natural denisties satisfy $d(A)+d(B)=1$ and we have $A\cap C \subset B$. Hence, using the PNT in aritmetic progressions, which tells that $d(C)=\frac{1}{2}$, we get $d(B)\geq d(A)-\frac{1}{2}=\frac{1}{2}-d(B)$, and from this we solve $d(B)\geq \frac{1}{4}$.
Answer to question 3: To clarify the definition, $C_n=\{O_n\pm d: d<p_{n+2} \hspace{0.1 cm}\text{is a power of two}\}.$ So the number of elements up to $x$ in $\bigcup_{n\geq 1} C_n$ grows like $\log^2 x$. As far as I know, no "natural" set that contains logarithmically few elements has been proved to contain infinitely many primes, so estimating the number of primes in $C_n$ seems hopeless. But one would expect that even $O_n+2$ is prime infinitely often.
Answer to question 4: first notice that the greatest common divisor of any two elements of $C_n$ is one. Indeed, if $q>2$ is a prime satisfying $q\mid O_n\pm 2^{\alpha}, q\mid O_n\pm 2^{\beta}$ and $2^{\alpha}\leq 2^{\beta}<p_{n+2}$, then $q\mid 2^{\beta-\alpha}-1,$ but this is a contradiction since $q\geq p_{n+2}$. So the question of $Q_n$ being divisible by an odd prime square reduces to an element of $C_n$ having this property. Again, this seems extremely hard since for example the following problem seems easier but is known to be open: Is $2^{p}-1$ ($p$ prime) always squarefree?
Answer to question 5: This would at least follow from Dickson's conjecture. According to it, there should even exist translates of $C_n$ that consist entirely (except $O_n\pm 1$) of primes. The constellation is admissible for this conjecture since if $q>2$ was a prime that divided one element of $kO_n+C_n$ for every $k$, then $q$ would be coprime to $O_n$ and bigger than $p_{n+1}$, but then $kO_n \pmod q$ would attain all the values $1,...,q-1$ as $k$ varies, which is a contradiction.