Let $(X_n)_{n\geq 1}$ be a sequence of i.i.d. standard normal variables. Let $(S_n)_{n\geq 0}$ and $(T_n)_{n\geq 0}$ be given by
$$ S_n = \sum_{i=1}^{n} X_i \qquad\text{and}\qquad T_n = \sum_{i=1}^{n} (X_i^2 - 1). $$
We will also fix a partition $\Pi = \{0 = t_0 < t_1 < \cdots < t_k = 1\}$ of $[0, 1]$. Then define
$$ \begin{gathered}
Y_n = \frac{1}{n}\sum_{i=1}^{n} | S_{i-1} | (X_i^2-1), \\
Y_{\Pi,n} = \frac{1}{n} \sum_{j=1}^{k} |S_{\lfloor nt_{j-1}\rfloor}| (T_{\lfloor nt_j\rfloor} - T_{\lfloor nt_{j-1} \rfloor}).
\end{gathered}$$
Ingredient 1. If $\varphi_{X}(\xi) = \mathbb{E}[\exp(i\xi X)]$ denotes the characteristic function of the random variable $X$, then the inequality $|e^{ix} - e^{iy}| \leq |x - y|$ followed by Jensen's inequality gives
\begin{align*}
\big| \varphi_{Y_n}(\xi) - \varphi_{Y_{\Pi,n}}(\xi) \big|^2
&\leq \xi^2 \mathbb{E}\big[ (Y_n - Y_{\Pi,n})^2 \big] \\
&= \frac{\xi^2}{n^2}\sum_{j=1}^{k} \sum_{i \in (nt_{j-1}, nt_j]} 2 \mathbb{E} \big[ \big( | S_{\lfloor n t_{j-1} \rfloor} | - | S_{i-1} | \big)^2 \big].
\end{align*}
From the reverse triangle inequality, the inner expectation is bounded by
\begin{align*}
2 \mathbb{E} \big[ \big( | S_{\lfloor n t_{j-1} \rfloor} | - | S_{i-1} | \big)^2 \big]
\leq 2 \mathbb{E} \big[ \big( S_{i-1} - S_{\lfloor n t_{j-1} \rfloor} \big)^2 \big]
= 2(i-1-\lfloor nt_{j-1} \rfloor),
\end{align*}
and summing this bound over all $i \in (nt_{j-1}, nt_j]$ yields
$$ \big| \varphi_{Y_n}(\xi) - \varphi_{Y_{\Pi,n}}(\xi) \big|^2
\leq \frac{\xi^2}{n^2} \sum_{j=1}^{k} (\lfloor n t_j \rfloor - \lfloor n t_{j-1} \rfloor)^2
\xrightarrow[n\to\infty]{} \xi^2 \sum_{j=1}^{k} (t_j - t_{j-1})^2. \tag{1} $$
Ingredient 2. From the Multivariate CLT, we know that
$$
\Bigg( \frac{S_{\lfloor nt_j\rfloor} - S_{\lfloor nt_{j-1}\rfloor}}{\sqrt{n}}, \frac{T_{\lfloor nt_j\rfloor} - T_{\lfloor nt_{j-1}\rfloor}}{\sqrt{n}} : 1 \leq j \leq k \Bigg)
\xrightarrow[n\to\infty]{\text{law}} ( W_{t_j} - W_{t_{j-1}}, N_j : 1 \leq j \leq k ),
$$
where $W$ is the standard Brownian motion, $N_j \sim \mathcal{N}(0, 2(t_j - t_{j-1}))$ for each $1 \leq j \leq k$, and all of $W, N_1, \cdots, N_k$ are independent. By the continuous mapping theorem, this shows that
$$ Y_{\Pi,n} \xrightarrow[n\to\infty]{\text{law}} \sum_{j=1}^{k} W_{t_{j-1}} N_j. $$
Moreover, conditioned on $W$, the right-hand side has normal distribution with mean zero and variance $2\sum_{j=1}^{k} W_{t_{j-1}}^2 (t_j - t_{j-1}) $, and so,
$$ \lim_{n\to\infty} \varphi_{Y_{\Pi,n}}(\xi) = \mathbb{E}\left[ \exp\bigg( -\xi^2 \sum_{j=1}^{k} W_{t_{j-1}}^2 (t_j - t_{j-1}) \bigg) \right]. \tag{2} $$
Ingredient 3. Again let $W$ be the standard Brownian motion. Since the sample path $t \mapsto W_t$ is a.s.-continuous, we know that
$$ \sum_{j=1}^{k} W_{t_{j-1}}^2 (t_j - t_{j-1}) \longrightarrow \int_{0}^{1} W_t^2 \, \mathrm{d}t $$
almost surely along any sequence of partitions $(\Pi_k)_{k\geq 1}$ such that $\|\Pi_k\| \to 0$. So, by the bounded convergence theorem,
$$ \mathbb{E}\left[ \exp\bigg( -\xi^2 \sum_{j=1}^{k} W_{t_{j-1}}^2 (t_j - t_{j-1}) \bigg) \right]
\longrightarrow \mathbb{E}\left[ \exp\bigg( -\xi^2 \int_{0}^{1} W_t^2 \, \mathrm{d}t \bigg) \right] \tag{3} $$
as $k\to\infty$ along $(\Pi_k)_{k\geq 1}$.
Conclusion. Combining $\text{(1)–(3)}$ and letting $\|\Pi\| \to 0$ proves that
$$ \lim_{n\to\infty} \varphi_{Y_n}(\xi) = \mathbb{E}\left[ \exp\bigg( -\xi^2 \int_{0}^{1} W^2_t \, \mathrm{d}t \bigg) \right], $$
and therefore $Y_n$ converges in distribution to $\mathcal{N}\big( 0, 2\int_{0}^{1} W_t^2 \, \mathrm{d}t \big)$ as desired.
Best Answer
Call the numerator $A_n$ and the denominator $B_n$. By the multidimensional CLT $(A_n/\sqrt{n}, B_n/\sqrt{n})$ converges jointly in distribution to a mean zero 2D Normal random variable $(A,B)$ with Var$(A) = 1$, Var(B) = 15 and Cov(A,B) = $3$ (consider the 2D vectors $(X_i,X_i^3)$ as your sample). You can then apply the continuous mapping theorem to see that $A_n/B_n$ converges to $A/B$. The distribution of this random variable can be computed by hand using change-of-variables and it is generalized Cauchy distributed.