Probability for success of exhaustive search on PRF keys
Let $q$ be the number of queries made by the attacker.
Consider that the view of $\mathcal{A}$ in the experiment is equivalent to their view in a modified experiment, where the oracle is restricted to the domain $\{1,\dots,q\}$.

Now consider, that in this modified experiment, there are exactly $2^q$ functions of the form $R : \{1,\dots,q\} \to \{0,1\}$.
If we sample one of those functions uniformly at random, what is the chance that one of the keys of the PRF will "hit" this function? Well, we can't say exactly, but we can give an upper bound. There are $2^n$ keys, which means at most, if each each of those keys hits a different function, $2^n$ functions can be hit.
This mean the probability of one of the keys hitting our uniformly sampled function is upper bounded by $$\frac{\# \text{ of keys}}{\# \text{ of functions}} = \frac{2^n}{2^q} = 2^{n-q}.$$

This means, we have
$$\Pr[\mathcal{A}^{R(\cdot)}=1] \leq 2^{n-q}$$

and also (by definition)
$$\Pr[\mathcal{A}^{F_k(\cdot)}=1] = 1.$$

Therefore
$$\left|\Pr\left[\mathcal{A}^{F_k(\cdot)}=1\right]-\Pr\left[\mathcal{A}^{R(\cdot)}=1\right]\right|\geq 1-2^{n-q}.$$

Now we need to choose a value for $q$ that makes this attack successful. If you plug in you suggested value of $q=n$, you'll notice, that we get $$1-2^{n-n}=1-2^0=1-1=0,$$
which is unfortunate.
We notice that as soon as we choose something a bit larger, say $q=n+1$ , we have a successful attack by standard definitions of PRFs:
$$1-2^{n-(n+1)}=1-2^{-1}=1-\frac{1}{2}=\frac{1}{2}$$
which is non-negligible.

However, we like to go for attacks that are successful with overwhelming probability and we like to go for simple easily recognizable negligible functions.
So we choose $q=2n$ and get 
$$1-2^{n-2n}=1-2^{-n}$$
which is easy to read and readily recognizable as an overwhelming probability.