Theorem 15.6 in Analysis on Manifolds by Munkres states
For an open set $A \subset{R}^n$ there exists a sequence $\{O_j\}$ of
open sets such that $O_j \subset O_{j+1}$
and $A = \cup_{j=1}^\infty O_j$. The extended integral $\int_A f$
exists if and only if the sequence $\int_{O_j} |f|$ is bounded, in which case we
have
$$\lim_{j \to \infty} \int_{O_j} f = \int_A f$$
I will add that each $O_j$ can be chosen as the interior of a compact rectifiable set $\bar{O}_j$ and, since, $f$ is continuous and bounded on $\bar{O}_j$ the integrals $\int_{\bar{O}_j} f = \int_{O_j} f$ exist properly. (The existence of $\int_{O_j} f$ over the interior and equality to $\int_{\bar{O}_j} f$ follows from Theorem 13.6 in Munkres.)
I think I have found a way to prove this statement for the case that $E$ has $content \ 0$. For this problem we are given that $W = V \cup E$ where $W$ and $V$ are open and $E$ is of measure $0$. We can assume that $V \cap E = \emptyset$ (since this seems to be the right way to state this theorem), so that $V=W-E=W \cap E^c$. We are also given that $f$ is continuous on $W$ and the extended integral $\int_V f$ exists.
Take a sequence $O_1 \subset O_2 \subset \ldots$ of open sets such that $W = \cup_{j=1}^\infty O_j$. Since $O_j \subset W$, then $O_j=O_j \cap W$. Thus the sets $\hat{O}_j = O_j \cap E^c=(O_j \cap W) \cap E^c=O_j \cap (W \cap E^c)=O_j \cap V$ are also open and and nested with union $V = \cup_{j=1}^\infty \hat{O}_j$.
We know that if $A$ has content $0$ then it is $rectifiable$ (it is bounded and its boundary has measure $0$) and it has measure $0$. Thus it can be proven (by Theorem 11.3) that $\int_A f$ exists and equals $0$.
Now since $O_j \cap E \subset E$ and $E$ has content $0$, then $O_j \cap E$ has content $0$. It follows that $\int_{O_j \cap E} |f|=0$. Then by Corollary 13.4
$$\int_{O_j} |f| = \int_{\hat{O_j}} |f| + \int_{O_j \cap E} |f| = \int_{\hat{O_j}} |f|$$.
Thus the sequence $\int_{O_j} |f|$ is bounded and the extended integral $\int_W f$ exists. Furthermore, as shown for $|f|$,
$$\int_{O_j} f = \int_{\hat{O_j}} f,$$
and, it follows that,
$$\int_W f = \lim_{j \to \infty}\int_{O_j} f = \lim_{j \to \infty}\int_{\hat{O_j}} f = \int_V f$$
Best Answer
If you just want to split the $z_{i}-z_{j}$ and don't care if you complicate the $f(z_{1},\cdots ,z_{n})$, then the Fourier representation is the way to go. To simplify, I will sketch the method using real variables x, the extension to complex variables is tedious, but I think the principle holds.
Start from $\prod_{i<j}g^{2}(|x_{i}-x_{j}|)=\prod_{i\neq j}g(|x_{i}-x_{j}|)$,
Define $g(x)$such that $g(x)=|x|^{-\frac{1}{2}\gamma^{2}}\text{for }x\in[0,1]$ and $0$ elsewhere.
Define $g=\int dke^{ikx}\tilde{g}(k)$ (the Fourier representation) and suppose you can evaluate it in closed form (exercise for the reader).
The Integral $$ I=\int dx_{1}\cdots dx_{n}\prod_{i\neq j}g(x_{i}-x_{j})f(x_{1},\cdots,x_{n})$$ becomes $$I =\int dx_{1}\cdots dx_{n}\int \prod_{i\neq j} dk_{ij}\tilde{g}(k_{ij})e^{i\sum_{i\neq j}k_{ij}(-x_{i}+x_{j})}f(x_{1},\cdots,x_{n})$$
let us examine the exponent: $\sum_{i\neq j}k_{ij}(-x_{i}+x_{j})=\sum_{i}x_{i}\sum_{j}(k_{ji}-k_{ij})$
So, $$I=\int dx_{1}\cdots dx_{n}\int \prod_{i\neq j} dk_{ij}\tilde{g}(k_{ij})e^{i\sum x_{i}\sum_{j}(k_{ji}-k_{ij})}f(x_{1}\cdots x_{n}).$$ Notice that the x integration is actually a Fourier transform of $f$ (at least if we extend the function $f$ to be zero outside the $x$ integration range. $$I=\int\prod_{ij} dk_{ij}\tilde{g}(k_{ij})\tilde{f}(\sum_{p}(k_{p1}-k_{1p}),\cdots,\sum_{p}(k_{pn}-k_{np}))$$ which is as far as I can get without making more assumptions.
Whether this helps, or just moves the difficulty elsewhere is up to you.