A more conveniently written and slightly more general version of the desired result is as follows: $$E\Big|\sum_{i\in[n]}X_i\Big|\ge E|X_1|,$$ where the $X_i$'s are iid random variables (r.v.'s) with a finite mean.

Let $p:=P(X_i\ge0)$, $q:=1-p$, $A:=EX_i^+$, $B:=EX_i^-$. Without loss of generality, $0<p<1$. For $J\subseteq[n]$, let $$I_J:=1(X_j\ge0\ \forall j\in J,X_j<0\ \forall j\notin J).$$ Following the lines of the first part of the proof for $n=2$, we have $$\begin{aligned} E\Big|\sum_{i\in[n]}X_i\Big| &=\sum_{J\subseteq[n]}E\Big|\sum_{i\in[n]}X_i\Big|\,I_J \\ &=\sum_{J\subseteq[n]}E\Big|\sum_{i\in[n]}X_i\, I_j\Big| \\ &=\sum_{J\subseteq[n]}E\Big|\Big(\sum_{i\in J}X_i^+-\sum_{i\notin J}X_i^-\Big)\, I_J\Big| \\ &\ge\sum_{J\subseteq[n]}\Big|E\Big(\sum_{i\in J}X_i^+-\sum_{i\notin J}X_i^-\Big)\, I_j\Big| \\ &=\sum_{J\subseteq[n]}\big||J|A/p-(n-|J|)B/q\big|\,p^{|J|}q^{n-|J|} \\ &=\sum_{k=0}^n\big|kA/p-(n-k)B/q\big|\,p^{k}q^{n-k}\binom nk=:s_n(p,A,B). \end{aligned}$$ Note also that $$s_n(p,A,B)=E\big|AX/p-(n-X)B/q\big|,$$ where $X$ is a r.v. with the binomial distribution with parameters $n,p$.

Without loss of generality, $A+B=1$. It remains to show that $$s_n(p,A,1-A)\ge1$$ for all $A\in[0,1]$. Graphing suggests this is true.

Unfortunately, the second, "squaring" part of the proof for $n=2$ will not work for large enough $n$; that is, the inequality $$\sum_{k=0}^n\Big(\big(kA/p-(n-k)B/q\big)\,p^{k}q^{n-k}\binom nk\Big)^2\ge(A+B)^2$$ will not hold if $n$ is large.

A more conveniently written and slightly more general version of the desired result is as follows: $$E\Big|\sum_{i\in[n]}X_i\Big|\ge E|X_1|,$$ where the $X_i$'s are iid random variables (r.v.'s) with a finite mean.

Let $p:=P(X_i\ge0)$, $q:=1-p$, $A:=EX_i^+$, $B:=EX_i^-$. Without loss of generality, $0<p<1$. For $J\subseteq[n]$, let $$I_J:=1(X_j\ge0\ \forall j\in J,X_j<0\ \forall j\notin J).$$ Following the lines of the first part of the proof for $n=2$, we have $$\begin{aligned} E\Big|\sum_{i\in[n]}X_i\Big| &=\sum_{J\subseteq[n]}E\Big|\sum_{i\in[n]}X_i\Big|\,I_J \\ &=\sum_{J\subseteq[n]}E\Big|\sum_{i\in[n]}X_i\, I_j\Big| \\ &=\sum_{J\subseteq[n]}E\Big|\Big(\sum_{i\in J}X_i^+-\sum_{i\notin J}X_i^-\Big)\, I_J\Big| \\ &\ge\sum_{J\subseteq[n]}\Big|E\Big(\sum_{i\in J}X_i^+-\sum_{i\notin J}X_i^-\Big)\, I_j\Big| \\ &=\sum_{J\subseteq[n]}\big||J|A/p-(n-|J|)B/q\big|\,p^{|J|}q^{n-|J|} \\ &=\sum_{k=0}^n\big|kA/p-(n-k)B/q\big|\,p^{k}q^{n-k}\binom nk=:s_n(p,A,B). \end{aligned}$$ Note also that $$s_n(p,A,B)=E\big|AX/p-(n-X)B/q\big|,$$ where $X$ is a r.v. with the binomial distribution with parameters $n,p$.

Without loss of generality, $A+B=1$. It remains to show that $$s_n(p,A,1-A)\ge1$$ for all $A\in[0,1]$. Graphing suggests this is true.

Unfortunately, the second, "squaring" part of the proof for $n=2$ will not work even for $n=3$ and $p$ close enough to $1/2$; that is, the inequality $$\sum_{k=0}^n\Big(\big(kA/p-(n-k)B/q\big)\,p^{k}q^{n-k}\binom nk\Big)^2\ge(A+B)^2$$ will not hold for such $n,p$.