Note that $$\frac{1}{2} \log 3 - \log 2 < 0$$ and therefore $$\log \left( \prod_{j=1}^n X_j \right) = \sum_{j=1}^n \log (X_j) = n \bigg(\underbrace{\frac{1}{n} \sum_{j=1}^n \log(X_j)}_{\to \frac{1}{2} \log 3 - \log 2 <0} \bigg) \to - \infty$$ almost surely. Now the continuity of $\exp$ entails
$$\prod_{j=1}^n X_j = \exp \left[ \log \left( \prod_{j=1}^n X_j \right) \right] \to 0 \qquad \text{a.s.}$$
Note that since $(Y_n,\mathcal F_n^Y)$ is a martingale, then uniform integrability is equivalent with convergence in $L_1$. Moreover, if the limit exists (call it $Y$), then it must be the case that $\mathbb E[Y]=1$ since $\mathbb E[Y_n] = 1$ for every $n\in \mathbb N$. However, as you noticed, we know (since it is non-negative martingale) that $Y_n \to Y$ almost surely, for some $Y \in [0,\infty)$. We'll show that $Y = 0$ almost surely.
Note that there exist $\delta,\epsilon > 0$ such that $\mathbb P(|\xi_k-1| > \epsilon) =\delta >0 $
By that we have $\sum_{k=1}^\infty \mathbb P(|\xi_k - 1| > \epsilon) = \infty$, and by Borel Cantelli, there is set $\Omega_0$ of measure $1$ such that for every $\omega \in \Omega_0$ there exist subsequence $(k_m(\omega))_{m \in \mathbb N}$ such that $|\xi_{k_m(\omega)}(\omega) - 1 | > \epsilon$ for every $m \in \mathbb N$.
Now, we need to use some tool from analysis, note that if infinite product $\prod_{k} a_k$ of nonnegative numbers $a_k$ converges to some $a \in (0,\infty)$ then $\lim_{k \to \infty} a_k = 1$ (it is due to taking logarithm of that product and noticing that $\lim_{k \to \infty} \ln(a_k) = 0$ iff $\lim_{k \to \infty} a_k = 1)$
But we showed that in our case for $\omega \in \Omega_0$ (so almost surely) $\xi_k(\omega) \not \to 1 $, which means that $Y$ must be $0$ or $\infty$ almost surely (the product must diverge). However, by martingale convergence theorem, we know that limit is finite almost surely, so it must be that $Y=0$ a.s. Hence it cannot converge in $L_1$, since $\mathbb E[|Y_n - Y|] = \mathbb E[|Y_n|] = 1 \not \to 0$
Best Answer
Let $M_n(q):= q^n\sqrt{Y_n}$. If we can find $q>1$ such that $(M_n(q))$ is a martingale with respect to $(\mathcal F_n)$, then we will be done, as $M_n(q)$ will be a non-negative martingale bounded in $\mathbb L^1$.
To find $q$, we can compute $\mathbb E\left[M_n(q)\mid\mathcal F_{n-1}\right]$ by using the pull-out property and independence. In order to find $M_{n-1}(q)$, we need $q$ such that $q\mathbb E\left[\sqrt{X_n}\right]=1$.