I am looking for an argument that for a 3-dim. Brownian motion started at some $x$ we have \begin{equation} | B_t | \to \infty \text{ a.s.} \end{equation} as $t \to \infty$.
As pointed out by the first comment, this does not hold for dimension 1.
Edit: Strongly edited to make the statement true.
The only nice solution I know uses Itô Calculus. However, it works equally well in $\mathbb{R}^n$ for $n\geq 3$.
Note that, given $0<r<R,$ the function $$f_{r,R}(x)=\frac{r^{-\frac{n-2}{2}}-|x|^{-\frac{n-2}{2}}}{r^{-\frac{n-2}{2}}-R^{-\frac{n-2}{2}}}$$ is harmonic in the annular region $A(r,R):=B(0,R)\setminus \overline{B}(0,r)$ and is constantly equal to $1$ on $\partial B(0,R)$ and constantly equal to $0$ on $\partial B(0,r)$. \
Hence, if $r<1<R$ and $B$ is Brownian Motion started at $e_1=(1,0,0)$ and $\tau_{r,R}$ is the first hitting time of $\partial A(r,R)$, we get that $f_{r,R}(B_{t\wedge \tau_{r,R}})$ is a bounded martingale. Accordingly, by Optional Sampling, we see that $\mathbb{P}^{e_1}(|B_{\tau_{r,R}}|=R)=f_{r,R}(1)$.
Note now that $$ \sum_{k=1}^{\infty}\mathbb{P}^{e_1}(|B_{\tau_{2^{-k},2^k}}|=2^{-k})=\sum_{k=1}^{\infty} (1-f_{2^{-k},2^k}(1))=\sum_{k=1}^{\infty} \frac{1^{-\frac{n-2}{2}}-2^{-k\frac{n-2}{2}}}{2^{k\frac{n-2}{2}}-2^{-k\frac{n-2}{2}}}<\infty $$
Now, by the Strong Markov Property and Brownian Scaling, we get that $t\mapsto \frac{1}{|B_{\tau_{r,R}}|}B_{t+\tau_{r,R}}$ is a time-scaled Brownian Motion started on the $n$-sphere and thus, letting $\tau^{k}$ denote the first hitting time of $B_{\tau^{k-1}}A(2^{-k},2^k)$ we can apply the Borel-Cantelli Lemma (and rotation invariance) to get that $\mathbb{P}^{e_1}(|B_{\tau^k}|=|2^{-k} B_{\tau^{k-1}}|\; \textrm{i.o.})=0$. Accordingly, $|B_{\tau^k}|=2^k |B_{\tau^{k-1}}|$ eventually with probability one.
Fix such an event $\omega$ and $k_0$ large enough such that $|B_{\tau^k}(\omega)|=2^k |B_{\tau^{k-1}}(\omega)|$ for all $k\geq k_0$. Then, for any such $k$, $(B_t(\omega))_{t\geq \tau^{k}}$ can never return to $\partial(B(0, 2^{-k_0} B_{\tau_{k_0}}(\omega))),$ since the alternative would violate our assumption on $k$. And thus, we get that $(|B_{\tau^k}|=2^k |B_{\tau^{k-1}}|\; \textrm{evt.})\subseteq (\exists n; s.t.\; B\; \textrm{leaves} \; B(0,2^{-n}) \; \textrm{forever}):=\mathcal{E},$ so $\mathbb{P}(\mathcal{E})=1$. However, by the strong Markov property and scaling (note that the above also proves that $|B_t|$ attains arbitrarily large values almost surely), if $B$ returns to some ball around $0$ infinitely often, it will have to return to any ball around $0$ infinitely often. Thus, with probability $1$, Brownian Motion leaves any fixed ball around $0$ forever at some point. Note now that
$$(|B_t|\to \infty)=\bigcap_{n=1}^{\infty} (B \; \textrm{leaves} \; B(0,n)\; \textrm{evt.})$$
and the last set is a countable intersection of almost sure events, and hence, almost sure. This proves the desired.