do we at least have $Y_n=\Theta(\frac{\log n}{n})$ almost surely?
The answer to this is no. It is not even true that $Y_n=\Theta(\frac{h(n)}{n})$ almost surely (a.s.) for any (deterministic or random) function $h$ such that $h(n)\to\infty$ a.s. (as $n\to\infty$). Indeed, suppose the contrary: that $\liminf_n nY_n=\infty$ a.s. Then, by the Fatou lemma, $$ \infty=E\infty\le\liminf_n EnY_n =\liminf_n n\frac1{n+1}=1, $$ a contradiction. $\quad\Box$
Let us now show that $Y_n\not\sim f(n)$ a.s. for any deterministic positive function $f$.
To prove this, suppose the contrary: that $Y_n\sim f(n)$ a.s. for some deterministic positive function $f$. Take any positive sequence $(a_n)$ such that $\sum_n a_n=\infty$. Then $$\sum_n P(X_n<a_n/2)=\sum_n a_n/2=\infty.$$ So, by the Borel--Cantelli lemma, events $\{X_n<a_n/2\}$ a.s. occur infinitely often (i.o.), that is, for infinitely many values of $n$. Therefore and because $\{X_n<a_n/2\}\subseteq\{Y_n<a_n/2\}$, we see that events $\{Y_n<a_n/2\}$ a.s. occur i.o. Recalling now the assumption that $Y_n\sim f(n)$ a.s., we conclude that $f(n)\le a_n$ i.o. In particular, $f(n)\le \frac1{n\ln n}=o(1/n)$ i.o.
So, $$Z_n:=nY_n\sim nf(n)=o(1) \tag{1}\label{1}$$ a.s. for $n=n_k$ and $k\to\infty$, where $(n_k)$ is a strictly increasing deterministic sequence of positive integers. Also, $EZ_n^2<2$. So, for $n$ as above, $$\begin{aligned} 1\leftarrow EZ_n&=EZ_n\,1(Z_n<4)+EZ_n\,1(Z_n\ge4) \\ &\le EZ_n\,1(Z_n<4)+EZ_n^2/4 \\ &<EZ_n\,1(Z_n<4)+1/2\to1/2 \end{aligned}$$ by \eqref{1} and dominated convergence. So, we have a contradiction. $\quad\Box$