For any function $f: \{1,-1\}^n \rightarrow \{1,-1\}$, there is a unique multilinear polynomial $p \in \mathbb{R}[x_1,\dots, x_n]$ for which $p(x)=f(x)$ for all $x \in \{1,-1\}^n$ (see e.g. Lemma 4.1 here). I will call this the polynomial representation of $f$.
I would like to know, for each $k=0,\dots, n$, what the polynomial representation of the function $f_k$ is, where $f_k(x)=1$ if and only if the number of $+1$'s appearing in $x$ is equal to $k$.
In the absence of such a polynomial representation, I would like upper and lower bounds on the number of monomial terms appearing in the polynomial representation of each $f_k$.
Edit: Here is a proof that the polynomial representation of any such $f$ is unique (I thank Neal Young for bringing up this subtlety. This fact is obvious for $\{0,1\}$-valued functions, but takes slightly more work for $\{-1,1\}$-valued functions.)
It suffices to prove that whenever $\sum_{I \in \{0,1\}^n} \alpha_I x_1^{I_1}\cdots x_n^{I_n}=0$ for all $x \in \{-1,1\}^n$, it holds that $\alpha_I=0$ for all $I \in \{0,1\}^n$. We prove this by induction on $n$.
For $n=1$, if $\alpha_0+\alpha_1 x_1=0$ for $x_1=\pm 1$, then clearly $\alpha_0=\alpha_1=0$.
Now suppose $\sum_{I \in \{0,1\}^n} \alpha_I x_1^{I_1}\cdots x_n^{I_n}=0$ for all $x \in \{-1,1\}^n$. Then
$$ \sum_{\substack{I \in \{0,1\}^n\\I_n=0}} \alpha_I x_1^{I_1}\cdots x_{n-1}^{I_{n-1}}=0 $$ and $$-\sum_{\substack{I \in \{0,1\}^n\\I_n=1}} \alpha_I x_1^{I_1}\cdots x_{n-1}^{I_{n-1}}=0 $$ for all $x \in \{1,-1\}^{n-1}$, so by the induction hypothesis, $\alpha_I=0$ for all $I \in \{0,1\}^n$. $\square$