I'm assuming your norm is actually given by:$\newcommand{\vertiii}[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1
\right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}}$
$$\vertiii x = \sup_{N \in \mathbb{N}} \left\|\sum_{n=1}^N\alpha_n(x)x_n\right\|$$
where $\alpha_n(x)$ is the unique scalar coefficient with $x_n$ in the above representation of $x \in X$.
As you mentioned, we have:
$$\|x\| = \left\|\sum_{n=1}^\infty \alpha_n(x)x_n \right\| = \left\|\lim_{N\to\infty}\sum_{n=1}^N \alpha_n(x)x_n \right\|= \lim_{N\to\infty}\left\|\sum_{n=1}^N \alpha_n(x)x_n \right\| \le \sup_{N \in \mathbb{N}} \left\|\sum_{n=1}^N\alpha_n(x)x_n\right\| = \vertiii x $$
For the converse inequality, we will show that $X$ equipped with $\vertiii{\cdot}$ is a Banach space.
Let $(f_n)_{n=1}^\infty$ be a Cauchy sequence with respect to $\vertiii{\cdot}$. We have to show it converges with respect to $\vertiii{\cdot}$.
First notice that for all $n \in \mathbb{N}$ the sequence $\big(\alpha_n(f_j)\big)_{j=1}^\infty$ is a Cauchy sequence of scalars. Indeed, for any $i, j \in \mathbb{N}$ we have:
\begin{align}
\left|\alpha_n(f_i) - \alpha_n(f_j)\right|\|x_n\| &= \|\alpha_n(f_i - f_j)x_n\|\\
&= \left\|\sum_{k=1}^n \alpha_n(f_i - f_j)x_n - \sum_{k=1}^{n-1} \alpha_n(f_i - f_j)x_n\right\|\\
&\le \left\|\sum_{k=1}^n \alpha_n(f_i - f_j)x_n\right\| + \left\|\sum_{k=1}^{n-1} \alpha_n(f_i - f_j)x_n\right\|\\
&\le 2\vertiii{f_i - f_j} \xrightarrow{i, j \to \infty} 0
\end{align}
$\big(\alpha_n(f_j)\big)_{j=1}^\infty$ is a Cauchy sequence of scalars so it converges: $\alpha_n(f_j) \xrightarrow{j\to\infty} \beta_n$. The idea now is to show that
$$f_j = \sum_{n=1}^\infty \alpha_n(f_j)x_n \xrightarrow{j\to\infty} \sum_{n=1}^\infty\beta_nx_n$$
with respect to $\vertiii{\cdot}$.
Let $\varepsilon > 0$. Since $(f_n)_{n=1}^\infty$ is Cauchy, there exists $j_0 \in \mathbb{N}$ such that for all $i, j \ge j_0$ holds:
$$\vertiii{f_i - f_j} < \frac\varepsilon4$$
For $i, j \ge j_0$ and $N \in \mathbb{N}$ we have:
$$\left\|\sum_{n=1}^N \alpha_n(f_i)x_n - \sum_{n=1}^N \alpha_n(f_j)x_n\right\| = \left\|\sum_{n=1}^N \alpha_n(f_i - f_j)x_n\right\| \le \vertiii{f_i - f_j} < \frac\varepsilon4$$
Letting $j \to \infty$ here gives:
$$\left\|\sum_{n=1}^N \alpha_n(f_i)x_n - \sum_{n=1}^N \beta_nx_n\right\| \le \frac\varepsilon4$$
Furthermore, for any $M, N \in \mathbb{N}$ with $M > N$ we have:
$$\left\|\sum_{n=M+1}^N \alpha_n(f_i)x_n - \sum_{n=M+1}^N \beta_nx_n\right\| \le \left\|\sum_{n=1}^N \alpha_n(f_i)x_n - \sum_{n=1}^N \beta_nx_n\right\| + \left\|\sum_{n=1}^M \alpha_n(f_i)x_n - \sum_{n=1}^M \beta_nx_n\right\| \le \frac\varepsilon4 + \frac\varepsilon4 =\frac\varepsilon2$$
Now, since $\sum_{n=1}^\infty \alpha_n(f_{j_0})x_n$ converges, there exists $N_0 \in \mathbb{N}$ such that for all $M, N \in \mathbb{N}$ we have:
$$N > M \ge N_0 \implies \left\|\sum_{n=M+1}^N \alpha_n(f_{j_0})x_n\right\| \le \frac\varepsilon4$$
So for all $M, N \in \mathbb{N}$ with $N > M \ge N_0$ we have:
$$\left\|\sum_{n=M+1}^N \beta_nx_n\right\|\le \left\|\sum_{n=M+1}^N \alpha_n(f_{j_0})x_n\right\| + \left\|\sum_{n=M+1}^N \alpha_n(f_{j_0})x_n -
\sum_{n=M+1}^N \beta_nx_n\right\| \le \frac\varepsilon4 + \frac\varepsilon2 = \frac{3\varepsilon}4 < \varepsilon$$
So, $\displaystyle \left(\sum_{n=1}^N \beta_nx_n\right)_{N=1}^\infty$ is a Cauchy sequence in $(X, \|\cdot\|)$. We conclude that $\displaystyle \sum_{n=1}^\infty \beta_n x_n$ converges. Set $f = \displaystyle \sum_{n=1}^\infty \beta_n x_n$.
Because $\{x_n\}_{n\in\mathbb{N}}$ is a Schauder basis, the representation $f = \displaystyle \sum_{n=1}^\infty \beta_n x_n$ is unique so we in fact have $\beta_n = \alpha_n(f)$.
Finally, for $i \ge j_0$ we have:
$$\vertiii{f_i - f} = \sup_{N\in\mathbb{N}}\left\|\sum_{n=1}^N\alpha_n(f_i - f)x_n\right\| = \sup_{N\in\mathbb{N}}\left\|\sum_{n=1}^N\alpha_n(f_i)x_n - \sum_{n=1}^N\beta_nx_n\right\| \le \frac\varepsilon4 < \varepsilon$$
We conclude $f_j \xrightarrow{\vertiii{\cdot}} f$ so $(X, \vertiii{\cdot})$ is complete.
Recall the Bounded Inverse Theorem:
Let $X$, $Y$ be Banach spaces and $A : X \to Y$ a bounded invertible linear map. Then $A^{-1} : Y \to X$ is also bounded.
Consider the identity map $I : (X, \vertiii{\cdot}) \to (X, \|\cdot\|)$. Since $\|\cdot\| \le \vertiii{\cdot}$, we have that $I$ is bounded. Both $(X, \vertiii{\cdot})$ and $(X, \|\cdot\|)$ are Banach spaces so the identity in the reverse direction $I : (X, \|\cdot\|) \to (X, \vertiii{\cdot})$ is also bounded.
Therefore, there exists $C > 0$ such that
$$\vertiii x = \vertiii{Ix} \le C\|x\|$$
We conclude that $\|\cdot\|$ and $\vertiii{\cdot}$ are equivalent.
