Assuming you mean $f(X_1,\ldots,X_n) = |\sum_{i=1}^N X_i|^{-1}$, the case $n\geq 2$ is actually easier than $n=1$. You could use that there exists a constant $C>0$ such that for any $x\in\mathbb{R}^n$ the expectation $\mathbb{E}[ 1 / |X_1 + x| ] < C$. Then automatically also $\mathbb{E}[ 1 / |X_1 + \ldots + X_n| ] < C$ by taking $x$ to have the distribution of $X_2 + \ldots + X_N$.

Assuming you mean $f(X_1,\ldots,X_n) = |\sum_{i=1}^N X_i|^{-1}$, the case $n\geq 2$ is actually easier than $n=1$. You could use that there exists a constant $C>0$ such that for any $x\in\mathbb{R}^n$ the expectation $\mathbb{E}[ 1 / |X_1 + x| ] < C$. Then automatically also $\mathbb{E}[ 1 / |X_1 + \ldots + X_n| ] < C$ by taking $x$ to have the distribution of $X_2 + \ldots + X_N$.

Note that this argument does not work for $n=1$, because if $|x|=1$ then $\mathbb{E}[ 1 / |X_1 + x| ] = \infty$ due to a logarithmic divergence.