By dyadic decomposition, it suffices to obtain bounds on the quantities $$ \int_{|x-a_i| \sim R_i \forall i} \left[ \frac{V(x,a_1,\dots,a_n)}{(R_1 \cdots R_n)^{n+1}}\right]^{\frac{1}{n-1}}\ dx$$ which, when summed over dyadic powers of two $R_1,\dots,R_n > 0$, are bounded uniformly over $a_1,\dots,a_n$.
By permuting we may assume that $R_1 \leq \dots \leq R_n$. By rescaling we can also normalize the set $\{a_1,\dots,a_n\}$ to have diameter $1$. The triangle inequality then lets us assume $R_n \gtrsim 1$.
As implicitly observed by the OP, the volume $V(x,a_1,\dots,a_n)$ is of size at most $O(R_1 \cdots R_n)$ by multiplying all the lengths emenating from $x$. On the other hand, since $a_1,\dots,a_n$ lie in a diameter one set and $x$ is at distance $O(R_1)$ from $x$, we also have the bound $O(R_1)$ by the base-times-height formula. [Note that this is a significantly superior bound in the bad case identified by the OP (after applying the normalization used here), where the $a_i$ are all at bounded distance from the origin and $x$ is very far away, so that $R_1 \sim \dots \sim R_n \ggg 1$.] Finally, the condition $|x-a_1| \sim R_1$ restricts $x$ to a set of measure $O(R_1^n)$. Thus the above integral can be bounded by $$ \lesssim \left[\frac{\min(R_1 \cdots R_n, R_1)^2}{(R_1 \cdots R_n)^{n+1}}\right]^{\frac{1}{n-1}} R_1^n.$$ If $R_1 \leq 1$, we bound this by $$ \left[\frac{(R_1 \cdots R_n)^2}{(R_1 \cdots R_n)^{n+1}}\right]^{\frac{1}{n-1}} R_1^n = \prod_{j=1}^n \frac{R_1}{R_j} \leq \frac{R_1}{R_n}$$ which is summable to $O(1)$ since $R_1 \leq 1$ and $R_n \gtrsim 1$ (and there are only $O(\log^{n-2} (R_n/R_1))$ choices of the intermediate $R_2,\dots,R_{n-1}$ for any fixed $R_1,R_n$). If $R_1 > 1$, we instead bound by $$ \left[\frac{R_1^2}{(R_1 \cdots R_n)^{n+1}}\right]^{\frac{1}{n-1}} R_1^n \leq \left[\frac{R_1^2}{(R_1^{n-1} R_n)^{n+1}}\right]^{\frac{1}{n-1}} R_1^n$$ $$ = \frac{1}{R_1^{\frac{n-3}{n-1}}R_n^{\frac{n+1}{n-1}}} \leq \frac{1}{R_n^{\frac{n+1}{n-1}}}$$ and this is also summable to $O(1)$ (for each $R_n \gtrsim 1$ there are only $O(\log^{n-1} R_n))$ choices for $R_1,\dots,R_{n-1}$).
