I don't think the conditions in the OP imply integrability of $|X_1|^{1/n}$. Here is a solution based on Borel-Cantelli's theorem and a little corollary of it:
Lemma 1: Suppose $(Y_n:n\in\mathbb{N})$ is an i.i.d sequence of random variables.
$$\frac{Y_n}{n}\xrightarrow{n\rightarrow\infty}0\quad\text{a.s. iff}\quad E[|Y_1|]<\infty$$
A proof of Lemma 1 is given at the end of this posting.
Solution to the OP:
Throughout the remaining of this posting, $(X_n:n\in\mathbb{N})$ is an i.i.d. sequence.
Sufficiency:
First assume that
$E[\log(1+|X_1|)]<\infty$ and that $P[|X_1|=0]<1$. Lemma 1 implies that $C:=\big\{\frac{\log(1+|X_n|)}{n}\xrightarrow{n\rightarrow\infty}0\big\}$ has full measure.
Claim I: $Z<\infty$ on $C$. If $Z(\omega)=\infty$, then there is sequence $n_k(\omega)$ such that $|X_{n_k(\omega)}|>2^{n_k(\omega)}$. Then $\limsup_k\frac{\log(1+|X_{n_k(\omega)}(\omega)|)}{n_k(\omega)}\geq\log2$ and so, $\omega\notin C$.
Claim II: $Z=1$ a.s. As $P[|X_1|=0]<1$, there is $0<\varepsilon<1$ such that $P[|X_1|>\varepsilon]>0$. Thus, $\sum_nP[|X_n|>\varepsilon]=\infty$, and by the reverse Borel Cantelli, $A=\{|X_n|>\varepsilon\,\text{i.o.}\}$ has full measure. For $\omega\in A\cap C$, there is a sequence $n_k$ (depending on $\omega$ such that
$$\frac{\log\varepsilon}{n_k}\leq \frac{\log|X_{n_k}|}{n_k}\leq\frac{\log(1+|X_{n_k}|)}{n_k}$$
Letting $k\rightarrow\infty$ yields $\lim_k|X_{n_k}|^{1/n_k}=1$. Hence $Z(\omega)=\limsup_n|X_n(\omega)|^{1/n}\geq1$.
Suppose $Z(\omega)>1$ and let $m_k(\omega)$ be a sequence such that $Z(\omega)=\lim_k |X_{m_k(\omega)}(\omega)|^{1/m_k(\omega)}$. Let $0<\varepsilon'<\varepsilon$ so that $Z-\varepsilon'>1$.
Then, for all $k$ large enough
$$\frac{\log(1+(Z(\omega)-\varepsilon')^{m_k(\omega)})}{m_k(\omega)}<\frac{\log(1+|X_{m_k(\omega)}(\omega)|)}{m_k(\omega)} $$
Letting $k\rightarrow\infty$ yields $0<\log(Z(\omega)-\varepsilon')\leq 0$ which is a contradiction.
Claims I and II imply that $Z=1$ a.s.
Necessity:
Now assume that $B=\{Z=1\}$ has full measure. This clearly implies that that $P[|X_1|=0]<1$. Let $\omega\in B$. Then, for $\varepsilon>0$, there is $N$ such that for $n\geq N$, $|X_n|<(1+\varepsilon)^n$. From
$$0\leq \frac{\log(1+|X_n|)}{n}\leq\frac{\log(1+(1+\varepsilon)^n)}{n}\xrightarrow{n\rightarrow\infty}\log(1+\varepsilon)$$
it follows that $\lim_n\frac{\log(1+|X_n|)}{n}=0$. By Lemma 1, $E[\log(1+|X_1|)]<\infty$.
Proof of Lemma I: For any $\varepsilon>0$,
$$\varepsilon\sum_{n\geq1}P[|Y_1|>\varepsilon n]\leq E[|Y_1|]\leq \varepsilon\sum_{n\geq0}P[|Y_1|>n\varepsilon]$$
If $E[|Y_1|]<\infty$, then $\sum_nP[|Y_n|>\varepsilon n]=\sum_nP[|Y_1|>n\varepsilon]<\infty$ for any $\varepsilon>0$. Borel-Cantelli's implies that
$P[|Y_n|>\varepsilon n,\,\text{i.o.}]=0$ for any $\varepsilon>0$. Let
$A_\varepsilon=\{|Y_n|>\varepsilon n\quad\text{i.o.}\}$. Then $P\big[\bigcap_mA^c_{1/m}\big]=1$. Notice that
$\bigcap_mA^c_{1/m}=\{\frac{Y_n}{n}\xrightarrow{n\rightarrow\infty}0\}$.
Conversely, if $E[|Y_1|]=\infty$, then $\sum_nP[|Y_1|>n]=\sum_nP[|Y_n|>n]=\infty$. The reverse Borel-Cantelly theorem implies that $P[|Y_n|>n\,\text{i.o.}]=1$, which means that $P\big[\frac{Y_n}{n}\xrightarrow{n\rightarrow\infty}0\big]=0$.
Comment: If in the assumptions in the OP $\limsup_n$ is replaced by $\lim_n$, that is $\lim_n|X_n|^{1/n}=1$ a.s. and $P[|X_1|=0]<1$, then we have
$$\lim_n|X_n|^{1/n}=1\quad\text{a.s.}\qquad\text{iff}\qquad E[|\log(|X_1|)|]<\infty$$
I think you’re overthinking this a little. You’ve already done the hard part above! You just need to put it together.
As you observed, using BC2, you can establish:
$$\mathbb{P}[X_n \geq n \text{ infinitely often}] = 1$$
Now we simply play with our probability space. First, we observe:
$$\{ \omega : X_n \geq n \text{ infinitely often}\} = \{ \omega : \limsup_n \frac{X_n}{n} \geq 1\}$$
Second, that:
$$\{\omega : \lim_n \frac{X_n}{n} = 0 \} \subseteq \{ \omega : \limsup_n \frac{X_n}{n} < 1\} = \Omega \setminus \{ \omega : \limsup_n \frac{X_n}{n} \geq 1\} $$
It then easily follows from the second observation that:
$$\mathbb{P}[\lim_n \frac{X_n}{n} = 0] \leq 1 - \mathbb{P}[\limsup_n \frac{X_n}{n} \geq 1]$$
But we already showed that $$\mathbb{P}[\limsup_n \frac{X_n}{n} \geq 1] = \mathbb{P} [X_n \geq n \text{ infinitely often}] = 1$$
This proves the desired result.
Best Answer
This is easily proved using Borel-Cantelli Lemma and the following well known fact:
For a non-negative random variable $Y$ we have $EY <\infty$ iff $\sum P(Y>n) <\infty$.
Now let $\epsilon >0$. Taking $Y=\frac {|X_1|} \epsilon$ we see that $E|X_1|<\infty$ iff $\frac{E|X_1|}{\epsilon}<\infty$ iff $\sum P\left(\frac {|X_1|} {\epsilon} >n\right) <\infty$ iff $\sum P\left(\frac {|X_n|} n >\epsilon\right) <\infty$ iff $P\left( \frac {|X_n|} n >\epsilon\hspace{0.2cm} \text{ i.o.}\right)=0$. Can you finish the proof?