Let
- $f\in C^3(\mathbb R)$ be positive
- $g:=\ln f$
- $d\in\mathbb N$, $$p_d(x):=\prod_{i=1}^df(x_i)\;\;\;\text{for }x\in\mathbb R^d$$ and $\lambda^d$ denote the Lebesgue measure on $\mathcal B(\mathbb R^d)$
- $\ell>0$, $\sigma_d:=\ell d^{-\alpha}$ for some $\alpha\in[0,1]$ and $$Q_d(x,\;\cdot\;):=\mathcal N(x,\sigma_d^2I_d)\;\;\;\text{for }x\in\mathbb R^d$$
- $X$ be a $\mathbb R^d$-valued random variable on $(\Omega,\mathcal A,\operatorname P)$ with $$X_\ast\operatorname P=p_d\lambda^d$$
- $Y$ be a $\mathbb R^d$-valued random variable on $(\Omega,\mathcal A,\operatorname P)$ with $$\operatorname P\left[Y\in B\mid X\right]=Q_d(X,B)\;\;\;\text{almost surely for all }B\in\mathcal B(\mathbb R^d)\tag0$$
Note that, by $(0)$, $(X,Y)_\ast\operatorname P=X_\ast\operatorname P\otimes \:Q_d$ is the product of the distribution $X_\ast\operatorname P$ of $X$ under $\operatorname P$ and the Markov kernel $Q_d$. Moreover, there is a $\mathbb R^d$-valued random variable $Z$ on $(\Omega,\mathcal A,\operatorname P)$ with $Z_\ast\operatorname P=\mathcal N_d(0,I_d)$ and $Y=X+\sigma_dZ$. It's easy to see that $X$ and $Y-X$ are independent.
Assume $g'$ is Lipschitz continous (and hence $g''$ is bounded) and $g'''$ is bounded.
I want to show that $$\liminf_{d\to\infty}\operatorname E\left[1\wedge\prod_{i=1}^d\frac{f(Y_i)}{f(X_i)}\right]=0\tag1,$$ if $\alpha<1/2$ and $$\lim_{d\to\infty}\operatorname E\left[1\wedge\prod_{i=1}^d\frac{f(Y_i)}{f(X_i)}\right]=1\tag2,$$ if $\alpha>1/2$.
By Taylor's theorem (and a measurable selection theorem), $$g(Y_i)-g(X_i)=g'(X_i)(Y_i-X_i)+\frac12g''(X_i)(Y_i-X_i)^2+\frac16g'''(W_i)(Y_i-X_i)^3\tag3$$ for some $\mathbb R^d$-valued random variable $W$ with $W_i\in[X_i,Y_i]$ or $W_i\in[Y_i,X_i]$ for all $i\in\left\{1,\ldots,d\right\}$.
Note that $$\prod_{i=1}^d\frac{f(Y_i)}{f(X_i)}=\exp\sum_{i=1}^d(g(Y_i)-g(X_i)).\tag4$$ So, we may be able to conclude by showing that $$\liminf_{d\to\infty}\sum_{i=1}^d(g(Y_i)-g(X_i))=-\infty\tag5,$$ if $\alpha<1/2$ and $$\liminf_{d\to\infty}\sum_{i=1}^d(g(Y_i)-g(X_i))=0\tag6,$$ if $\alpha>1/2$, in suitable modes of convergence.
Maybe we can apply the central limit theorem to $$S_1^{(d)}:=\sum_{i=1}^dg'(X_i)(Y_i-X_i)$$ and the strong law of large numbers to $$S_2^{(d)}:=\sum_{i=1}^dg''(X_i)(Y_i-X_i)^2.$$
I'm not sure what we could do with $$S_3^{(d)}:=\sum_{i=1}^dg'''(W_i)(Y_i-X_i)^3.$$ The only thing I've noticed is that $$\operatorname E\left[\left|S_3^{(d)}\right|\right]\le2\sqrt{\frac2\pi}\left\|g'''\right\|_\infty d\sigma_d^3.\tag7$$ So, as long as $d\sigma_d^3\xrightarrow{d\to\infty}0$ (which should be the case precisely when $\alpha>1/3$), $S_3^{(d)}\xrightarrow{d\to\infty}0$ in $L^1$.