Using the fact that $(X_k)$ are identically distributed, we have
$$ \sum_{k=1}^{\infty} \Bbb{P}(A_k)
= \sum_{k=1}^{\infty} \Bbb{P}(X_1 > k \epsilon)
= \sum_{k=1}^{\infty} \Bbb{E}[\mathbf{1}_{\{ X_1 > k \epsilon \}}]
= \Bbb{E}\bigg[ \sum_{k=1}^{\infty} \mathbf{1}_{\{ X_1 > k \epsilon \}} \bigg]. \tag{*}$$
Now let us focus on the random variable $Y := \sum_{k=1}^{\infty} \mathbf{1}_{\{ X_1 > k \epsilon \}}$. It is not hard to see that
$$ Y = \begin{cases}
k, & X_1 \in (k\epsilon, (k+1)\epsilon] \text{ for some } k = 0, 1, \cdots \\
0, & X_1 = 0
\end{cases} $$
This proves that $Y \leq \frac{1}{\epsilon}X_1$ and hence
$$ \sum_{k=1}^{\infty} \Bbb{P}(A_k) = \Bbb{E}[Y] \leq \frac{1}{\epsilon}\Bbb{E}[X_1] < \infty. $$
Now all you have to do is to apply the 1st Borel-Cantelli's lemma.
Remark. If we replace the assumption $\Bbb{E}[X_1] < \infty$ by $\Bbb{E}[X_1] = \infty$, then a similar argument shows that $\limsup_{n\to\infty} \frac{X_n}{n} = \infty$ a.s. Indeed, one has the inequality
$$ Y \geq \frac{1}{\epsilon}X_1 - 1.$$
Plugging this back to the computation $\text{(*)}$, we have
$$ \sum_{k=1}^{\infty} \Bbb{P}(A_k) = \Bbb{E}[Y] \geq \frac{1}{\epsilon}\Bbb{E}[X_1] - 1 = \infty. $$
Therefore by the 2nd Borel-Cantelli lemma, $\Bbb{P}(\frac{X_n}{n} > \epsilon \text{ i.o.}) = 1$ and thus $\limsup_{n\to\infty} \frac{X_n}{n} \geq \epsilon$. Since $\epsilon > 0$ is arbitrary, the conclusion follows by letting $\epsilon \to \infty$ (along a subsequence, if you want to be rigorous).
Best Answer
Unfortunately, it seems hard to get a good control on $\mathbb{E}|T_n-\mu|$.
However, we can use the classical strong law of large numbers. First, using the fact that $\sum_{i=1}^ni^{-1}/\log n\to 1$, we can assume that $\mu=0$. Then denote $S_i=\sum_{k=1}^iX_i$, $S_0=0$ and write $$ \frac 1{\log n}\sum_{i=2}^n \frac{S_i-S_{i-1}}i=\frac 1{\log n}\sum_{i=2}^n \left(\frac{S_i}{i}-\frac{S_{i-1}}{i-1}\right)+\frac 1{\log n}\sum_{i=2}^n S_{i-1}\left( \frac{1}{i-1}-\frac 1i\right). $$ The first sum telescops, for the second one, use the fact that if $c_i\to 0$, then $$ \frac 1{\log n}\sum_{i=2}^nc_i/i\to 0. $$