A possible choice of space $\Omega$ to define Brownian motion is $\Omega=C(\mathbb R_+,\mathbb R)$, then the Brownian motion $(B_t)_{t\in\mathbb R_+}$ is simply the coordinate process, that is, for every $t$ in $\mathbb R_+$ and $\omega$ in $\Omega$, $B_t(\omega)=\omega(t)$.
In this construction, sample paths are the elements $\omega$ of $\Omega$.
But, as is usual in probability, one may prefer not to specify $\Omega$. Then $\Omega$ can be any space large enough for a family $(X_t)_{t\in\mathbb R_+}$ of random variables with the prescribed properties to exist on $\Omega$. Then a sample path is, for some $\omega$ in $\Omega$, the function $X(\omega):\mathbb R_+\to\mathbb R$, $t\mapsto X_t(\omega)$.
Lévy's construction by dichotomy, which you recall, might then correspond to $\Omega=S^\mathbb N$ the product of a countable number of copies of a probability space $(S,\mathcal S,Q)$ being large enough for one standard normal random variable $\xi$ to be defined on it. Then a Brownian motion $(X_t)_{t\in\mathbb R_+}$ on $\Omega$ can be defined as Lévy indicated using the i.i.d. copies of $\xi$ defined on each factor of $\Omega$. Thus, every $\omega$ in $\Omega$ is $\omega=(s_n)_{n\in\mathbb N}$ for some $s_n$ in $S$, and the random variables $X_1$ and $X_{1/2}$, say, are defined by $X_{1}(\omega)=\xi(s_1)$ and $X_{1/2}(\omega)=\frac12\xi(s_1)+\frac12\xi(s_2)$.
Edit: Recall that the $n$th approximation $X^{(n)}$ of the Brownian motion $X$ on $[0,1]$ is piecewise linear on each interval $[(k-1)/2^n,k/2^n]$ with $1\leqslant k\leqslant2^n$. Thus, the $0$th approximation is such that $X^{(0)}_t(\omega)=t\xi(s_1)$ for every $t$ in $[0,1]$, after that, $X^{(n+1)}_t=X^{(n)}_t$ at every $t=k/2^n$ and $2X^{(n+1)}_t=X^{(n)}_{k/2^n}+X^{(n)}_{(k+1)/2^n}+\xi(s_*)/2^{n/2}$ at every $t=(2k+1)/2^{n+1}$, where $*$ is the first index $i$ such that $\xi(s_i)$ is not used yet.
It actually is a form of CLT. Even though the $X_n$ are not independent, they are weakly dependent - $X_1$ and $X_n$ are very nearly independent, for large $n$. See weakly dependent CLT for example. At a glance, I found this paper which may interest you.
Best Answer
You can do but it is not so easy since the increments of fBm are dependence. In fact, it is well known that \begin{eqnarray}\label{Bht} B^{H}_{t}=\int^{t}_{0}z_{H}(t,s)dB_{s}\,, \end{eqnarray}where $B_{t}$ is a standard Brownian motion and the deterministic kernel is given by \begin{eqnarray}\label{zh} z_{H}(t,s)=c_{H}\left( \left(\frac{t}{s}\right)^{H-\frac{1}{2}}\left(t-s\right)^{H-\frac{1}{2}}- \left(H-\frac{1}{2}\right)s^{\frac{1}{2}-H}\int_{s}^{t}u^{H-\frac{3}{2}} \left(u-s\right)^{H-\frac{1}{2}}du\right){\bf 1}_{(0,t)}(s), \end{eqnarray} with the constant \begin{eqnarray}\label{ch} c_{H}^2 =\frac{2H\Gamma(\frac{3}{2}-H)}{\Gamma(H+\frac{1}{2})\Gamma(2-2H)}\;, \end{eqnarray} and ${\bf 1}_{(0,t)}(s)$ is an indicator function that takes the value 1 iff $s\in(0,t)$ and is zero otherwise.
Then Inspired by sottinen (2001, FS jounral), for $H\in(0,1)$, we define: \begin{equation}\label{Bthn} B_{t}^{H,n}=\sum\nolimits_{i=1}^{\lfloor nt\rfloor} \sqrt{n} (\int_{\frac{i-1}{n}}^{\frac{i}{n}} z_{H}(\frac{\lfloor nt\rfloor}{n},s)ds)\xi_{i} \;, \end{equation} where $\xi_{1},\ldots,\xi_{n}$ is an $n$-tuple of independent symmetric Bernoulli random variables with $\mathbb{P}(\xi_{i}=\pm 1)=\frac{1}{2}$, living on a probability space $(\Omega_{n}, \mathcal {F}_{n}, \mathbb{P}_{n})$. Now you can similate fbm using $B_{t}^{H,n}$.