Suppose $S$ is smooth $n-1$-dimensional closed and bounded (compact) hypersurface in $\mathbb{R}^n$. Suppose for simplicity that $S$ is the boundary of a Lipschitz domain, for example. Whatever makes calculations work.
Let $f:S \to \mathbb{R}$ be a function and $p < n-1$.
We have the integral $$I=\int_{S}\frac{f(x)}{|x|^{p}}d\sigma(x)$$ where $\sigma(x)$ is the surface measure/density on $S$, and I want to bound this as $$I \leq C\lVert f \rVert_{L^1(S)}$$ where $C$ is a constant that depends on $n-1$ and the size of the domain $S$ in some way.
How can I do show this? For the case where $S$ is just an open set, we can use polar coordinates (see this thread: Bounding an integral $\int_\Omega \frac{1}{|x-y|^{q}}\;dx$)
Is it possible to do this bound via polar coordinates then? I am confused with the surface measure part and how that will change.
Based on the information given, the best you can do is $$ \int_{S}\frac{f(x)}{|x|^{p}}d\sigma(x) \le \frac{1}{d^p} \|f\|_{L^1(S)} \tag{1}$$ where $d=\operatorname{dist}(0,\operatorname{supp}f)$. Note that $d\ge \operatorname{dist}(0,S)$, so you can use the latter distance, if the distance to the support cannot be estimated otherwise. If the support of $f$ contains $0$, you have no control of the integral in terms of the $L^1$ norm of $f$.
The proof of (1) is trivial: $|x|^{-p}\le d^{p}$ for all $x$ that contribute to the integral, hence $$ \int_{S}\frac{f(x)}{|x|^{p}}d\sigma(x) \le \int_{S} \frac{|f(x)|}{d^{p}}d\sigma(x) $$
To see that (1) cannot be improved, take $f$ supported in some shell $\{x:d\le |x|\le d+\epsilon\}$. In this case $$ \int_{S}\frac{f(x)}{|x|^{p}}d\sigma(x) \ge \int_{S} \frac{|f(x)|}{(d+\epsilon)^{p}}d\sigma(x) $$