Given $f:\{0,1\}^n \rightarrow \{-1,1\}$, find a subcube with large volume and large average value
Here is a better bound on the sample complexity. (Although the computational complexity is still $n^k$.)

Theorem. Assume there exists a subcube $S$ of size $2^{n-k}$ such that $|\mathbb{E}_{x \in S}[f(x)]| \geq 0.12$. With $O(2^k \cdot k \cdot \log n)$ samples we can, with high probability, identify a subcube $S'$ of size $2^{n-k}$ such that $|\mathbb{E}_{x \in S'}[f(x)]| \geq 0.1$.

Note the small loss in parameters ($0.12$ is optimal versus guarantee of $0.1$). 

Proof. Pick $m$ points $P \subset \{0,1\}^n$ uniformly at random and query $f$ at each $x \in P$.

Fix a subcube $S$ of size $2^{n-k}$. We have $\mathbb{E}[|S \cap P|]=m 2^{-k}$. By a Chernoff bound, $$\mathbb{P}[|S \cap P| &lt; m 2^{-k-1}] \leq 2^{-\Omega(m 2^{-k})}.$$ Also $$\mathbb{P}[| \mathbb{E}_{x \in S \cap P}[f(x)] - \mathbb{E}_{x \in S}[f(x)] | &gt; \varepsilon] \leq 2^{-\Omega(|S \cap P| \varepsilon^2)}.$$

By a union bound over all ${n \choose k}2^k$ choices of $S$, we have $$\mathbb{P}[\forall S ~~ | \mathbb{E}_{x \in S \cap P}[f(x)] - \mathbb{E}_{x \in S}[f(x)] | \leq \varepsilon] \geq 1 - {n \choose k} 2^k 2^{-\Omega(m 2^{-k} \varepsilon^2)}.$$ So by picking $m = O(2^k/\varepsilon^2 \cdot k \log n)$, we can ensure that, with probability at least $0.99$, we can estimate $\mathbb{E}_{x \in S}[f(x)]$ to within $\varepsilon$ for all subcubes $S$ of size $2^{n-k}$.

Setting $\varepsilon=0.01$, we prove the theorem: picking the subcube with the largest $| \mathbb{E}_{x \in S \cap P}[f(x)]|$ will, with high probability, satisfy the requirements. Q.E.D.