The function $f(n)$ must be $\Theta (\log n)$. **Update:** As Didier Piau points out in the comments, we can say something stronger: $\frac{f(n)}{\log_2 n} \to L$ as $n \to \infty$. See the update at the end of the argument.

Suppose, for some positive $L$ (the negative case is similar), $$\lim_{n \to \infty} \left(f(n) - 2^{-n} \sum_{k=0}^n \binom{n}{k} f(k) \right) = L.$$
Thus $$f(n) = L + r(n) + 2^{-n} \sum_{k=0}^n \binom{n}{k} f(k)$$ for some function $r(n)$ such that $r(n) \to 0$ as $n \to \infty$. This gives us a recurrence relation for $f(n)$. Since $r(n) \to 0$, $L + r(n) > 0$ for all sufficiently large $n$. So, for all sufficiently large $n$, there exist positive constants $C$ and $D$ such that $C < L + r(n) < D$. Since the initial terms in the function eventually become negligible in determining the value of $f(n)$ via the recurrence relation, there exist functions $g(n)$ and $h(n)$ such that for all sufficiently large $n$, $g(n) \leq f(n) \leq h(n)$ and $g(n)$ and $h(n)$ satisfy
$$g(n) = C + 2^{-n} \sum_{k=0}^n \binom{n}{k} g(k),$$
$$h(n) = D + 2^{-n} \sum_{k=0}^n \binom{n}{k} h(k).$$

So the problem reduces to showing that $g(n)$ and $h(n)$ are $\Theta (\log n)$. The argument is the same for both.

There are some different ways to do this; my favorite is to interpret the $g(n)$ recurrence probabilistically. Suppose we start at time $g(0)$, we flip a set of $n$ coins simultaneously each round, and it takes $C$ time units to do one round of flips. When a coin turns up heads for the first time, we cease flipping it. Let $T(n)$ be the time at which the last coin to achieve its first head does so. To find $E[T(n)]$, condition on the number of coins that achieve heads in the first round of flips. This yields $$E[T(n)] = C + 2^{-n} \sum_{k=0}^n \binom{n}{k} E[T(n-k)] = C + 2^{-n} \sum_{k=0}^n \binom{n}{k} E[T(k)].$$ Thus $g(n) = E[T(n)]$.

Another way to view $T(n)$ is that it is $g(0) + C M_n$, where $M_n = \max\{X_1, X_2, \ldots, X_n\}$ and the $X_i$'s are independent and identically distributed geometric $(1/2)$ random variables. (Each geometric random variable models the first time a head appears.) Thus $g(n) = g(0) + C E[M_n]$. It is known that $\frac{H_n}{\log 2} \leq E[M_n] \leq \frac{H_n}{\log 2} + 1$ and, more precisely, that $E[M_n]$ is logarithmically summable to $\frac{H_n}{\log 2} + \frac{1}{2}$. (See, for example, Bennett Eisenberg's paper "On the expectation of the maximum of IID geometric random variables," *Statistics and Probability Letters* 78 (2008) 135-143. See also this MO question, "What is the Expected Maximum out of a Sample (size $n$) from a Geometric Distribution?")

Thus $g(n) = \frac{C}{\log 2} \log n + O(1)$, which means that $h(n) = \frac{D}{\log 2} \log n + O(1)$ and $f(n) = \Theta (\log n)$.

**Update:** Given $\epsilon > 0$, if we take take $C > 0$ such that $L - \epsilon \leq C < L$ and $D = L + \epsilon$, this argument shows that
$$L - \epsilon + O\left(\frac{1}{\log n}\right) \leq \frac{f(n)}{\log_2 n} \leq L + \epsilon + O\left(\frac{1}{\log n}\right).$$

Thus, as $n \to \infty$, $\frac{f(n)}{\log_2 n} \to L.$

For some other ideas that pertain to this result, including what are effectively some alternative derivations, see Pradipta's recent MO question, "Coin Flipping and a Recurrence Relation". In fact, reading Pradipta's question and some of its answers gave me the ideas needed to construct this argument! So, thanks go to Pradipta, Didier Piau, Emil Jeřábek, and Louigi Addario-Berry.