Suppose $f_n \rightharpoonup f$ in $L^p(\mathbb{R})$, the boundedness of $f_n$ is an immediate consequence of the following more general statement: let $(E,\| \cdot \|_{E})$ be a normed vector space over $\mathbb{R}$ and $(x_n)_n \subset E,x \in E$, then
$$ x_n \rightharpoonup x \implies \sup_{n} \| x_n \|_{E} < +\infty $$
i.e, every weakly convergent sequence is bounded.
The proof is classical and relies on the canonical map properties (in particular, isometry), the uniform boundedness principle and the dual characterization of the norm.
Suppose $p \in (1,+\infty)$. As usual, if $f \in L^p$, we identify $f$ with its induced distribution $T_f : \mathcal{D}(\mathbb{R}) \mapsto \mathbb{R}$ such that $T_f(\phi)=\int_{\mathbb{R}} f\phi$ for every $\phi \in \mathcal{D}(\mathbb{R})$.
Recall that $f_n \rightharpoonup f$ in $L^p(\mathbb{R})$ is equivalent to say:
$$ \int_{\mathbb{R}} f_n g \rightarrow \int_{\mathbb{R}} fg \qquad \forall g \in L^{p'}(\mathbb{R}) $$
On the other hand, $C^{\infty}_c (\mathbb{R}) \subset L^{p'}$, whence $f_n \rightarrow f $ in the sense of distributions.
We now prove the reverse implication. Again suppose $p \in (1,+\infty)$. Let $c=\max\{ \sup_n \| f_n \|_{p}, \| f \|_{p} \}<\infty $ and $p'$ be Holder conjugate of $p$. Let $g \in L^{p'}$, we want to prove that
$$ \int_{\mathbb{R}} (f_n-f)g \rightarrow 0 $$
Since $ (C_c^{\infty}(\mathbb{R}), \| \cdot \|_{p'})$ is dense in ($L^{p'},\| \cdot \|_{p'})$, for every $\epsilon > 0$ there exists a function $\rho \in C_c^{\infty}(\mathbb{R}) $ such that $ \|\rho-g\|_{p'} \le \frac{\epsilon}{2c}$ .
For every $ n \in \mathbb{N}$ it holds:
\begin{align*} \| (f_n-f)g \|_{1} = \| (f_n-f)(g-\rho+\rho) \|_{1} \le \|(f_n-f)(g-\rho) \|_{1}+\| (f_n-f) \rho \|_{1} \\ \le \| (f_n-f) \rho \|_{1} + \| f_n - f \|_{p} \| g-\rho \|_{p'} \le \| (f_n-f) \rho \|_{1} + 2c \| g-\rho \|_{p'} \\ \le \| (f_n-f) \rho \|_{1} + \epsilon \end{align*}
Where we both used Holder's and Minkowski's inequalities.
Since $\epsilon$ was arbitrary and $f_n \rightarrow f$ on test functions, the claim follows taking the limit as $ n \rightarrow +\infty $.
To conclude, let's see a counterexample for the right-left implication if $p=1$. Let
$$ f_n(x)=\chi_{[n,n+1]} (x) \qquad x \in \mathbb{R} $$
We have $\sup_{n} \| f_n \|_{1} = 1 $ and $f_n \rightarrow 0 $ in the sense of distributions because for sufficiently large values of $n$, $f_n$ escapes the compact support of every test function. On the other hand the weak convergence in $L^1$ to the zero function would imply:
$$ \int_{\mathbb{R}} f_n g \rightarrow 0 \qquad \forall g \in L^{\infty}(\mathbb{R}) $$
Take $g=1$ to see that this is impossible. It's worth to note that the above proof in the case $p=1$ fails because test functions are not dense in $L^{\infty}$, hence the function $\rho$ that approximates $g$ may not exist.
As a final remark, note that "$ \leftarrow$" holds if $p=\infty$ replacing weak convergence with weak * convergence. Indeed, if $p \in (1,+\infty) $, the statement still trivially holds true in both directions since in these hypothesis $(L^p(\mathbb{R}, \| \cdot \|_{p})$ is reflexive and weak * convergence and weak convergence coincide in reflexive spaces.
Best Answer
As you noticed, it suffices to show that the sequence is uniformly integrable. There are several equivalent formulations of this. Since (as you noted) the sequence $(f_n)_{n \in \Bbb{N}}$ is bounded in $L^1$, it suffices to prove that $\sup_n \int_0^1 |f_n(x)| \cdot 1_{|f_n(x)| \geq M} \, d x \to 0$ as $M \to \infty$.
That this is indeed satisfied can be verified as follows: \begin{align*} & \int_0^1 |f_n (x)| \cdot 1_{|f_n(x)| \geq M} \, d x \\ & = \frac{1}{n} \int_0^1 n \cdot |f(n x)| \cdot 1_{|f(nx)| \geq M} \, d x \\ & = \frac{1}{n} \int_0^n |f(y)| \cdot 1_{|f(y)| \geq M} \, d y \\ & = \frac{1}{n} \sum_{i=0}^{n-1} \int_0^1 |f(y+i)| \cdot 1_{|f(y+i)| \geq M} \, d y \\ & \overset{(\ast)}{=} \frac{1}{n} \sum_{i=0}^{n-1} \int_0^1 |f(z)| \cdot 1_{|f(z)| \geq M} \, d z \\ & = \int_0^1 |f(z)| \cdot 1_{|f(z)| \geq M} \, d z. \end{align*} Here, we used the periodicity of $f$ at the step marked with $(\ast)$.
Note that the right-hand side of the above estimate is independent of $n$, and converges to zero as $M \to \infty$.