>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. $\quad\Box$

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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. $\quad\Box$ 

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In fact, $Y_n\not\sim f(n)$ even in probability for any positive deterministic  function $f$. 

Indeed, suppose that $Y_n\sim f(n)$ in probability for some positive deterministic function $f$. Then, by the Fatou lemma, 
$$1=E\lim_n\frac{Y_n}{f(n)}\le \liminf_n\frac{EY_n}{f(n)}
=\liminf_n\frac{1}{(n+1)f(n)},$$
so that $f(n)\lesssim1/n$. So, $nf(n)\to c$ for some real $c\ge0$, $n=n_k$, and $k\to\infty$, where $(n_k)$ is some strictly increasing sequence of positive integers. So, for such $n$ and $Z_n$ as above,
$$Z_n\to c$$
in probability. Also, $EZ_n\to1$, $EZ_n^2\to2$, and $EZ_n^4\to24$. Take now any real $A>0$. Then 
$$E(Z_n-c)^2=E(Z_n^2-2cZ_n+c^2)\to C:=2-2c+c^2,$$ 
whereas, for $n$ as above, 
$$\begin{aligned}
E(Z_n-c)^2
&=E(Z_n-c)^2\,1((Z_n-c)^2\le A)+E(Z_n-c)^2\,1((Z_n-c)^2>A) \\ 
&\le E(Z_n-c)^2\,1((Z_n-c)^2\le A)+E(Z_n-c)^4/A \\ 
&\le E(Z_n-c)^2\,1((Z_n-c)^2\le A)+2^3(EZ_n^4+c^4)/A \\ 
&\to 0+2^3(24+c^4)/A
\end{aligned}$$ 
by dominated convergence. We conclude that $0<C\le 2^3(24+c^4)/A$ for all real $A>0$, a contradiction. $\quad\Box$