In other words, given the set $[n] = \{1,2,\ldots,n\}$, we must find the least $k$ so that, for any family $\mathcal{F}$ of subsets of $[n]$ (elements of $\mathcal{P} = \mathcal{P}([n])$) with $|\mathcal{F}| = k$, there exists a mapping $\phi \colon \mathcal{F} \to \mathcal{P}$ such that $\phi(F) \in \{F, \overline{F}\}$ for all $F \in \mathcal{F}$ and $\displaystyle \bigcup_{F \in \mathcal{F}} \phi(F) = [n]$.

The answer is $k = \lceil \log_2 (n+1) \rceil$. The trick is a simple greedy solution. For each paper from the left to the right, Merlin takes the side with more numbers that aren't visible on any of the pages to the left. This guarantees that the number of unseen integers halves after every paper. For example, for $n = 7$, the pages are $(12356, 47), (1245, 367), (146, 2357)$. Before any page is inspected, Merlin hasn't seen any number. Now, to the first paper, Merlin should choose the one with more unseen numbers. The front side, $12356$, has more unseen numbers, so he takes that. This leaves $47$. The second paper, both sides have one number (the front side has $4$ and the back side has $7$), so Merlin can choose either. Let's suppose he takes the front side, leaving $7$ unseen. The third paper, clearly he should turn the paper so he sees the $7$, and his task is complete. To prove that this is minimal, Daniel simply has to make sure to split the unseen numbers as evenly as possible, so Merlin can do no better than halving the number of unseen integers. I haven't found an explicit construction for this, though.

Think this works for a construction. We only need to show that for $2^k$ it is at least $k+1$ pages. To make it easier we pretend that $2^k$ is $0$. Then change all numbers into $k$ digit binary form. The numbers written on front of page $i$ are all the numbers with $1$ as $i$th binary digit from the left, for $i=1$ to $k$. Then for needed $k+1$, write numbers with sum of binary digits even on front page. If we don't use all the pages, then we can assume that we use all but $1$. Let's say page $j$, $j<k+1$ was not needed. Let $a_i$ be opposite parity of the number chosen on page $i$ ($0,1$). So numbers with binary digits $a_1a_2...a_{i-1}1a_{i+1}...a_k$ and $a_1a_2...a_{i-1}0a_{i+1}...a_k$ are not chosen yet, and they do not appear on same side of page $k+1$, contradiction. If $j=k+1$, then $a_1a_2....a_k$ is not chosen, also contradiction. Hence at least $k+1$ pages are needed.