Let $X$ be the completed space with your norm.
Using the general fact that $|x|_p \leq |x|_q$ if $p \geq q$, your norm dominates the $\ell^2$ norm on sequences, and hence the completion of your space $X$ embeds into $\ell^2$.
You have $\ell^1$ is dense in $X$ (defined as the completion) using the $|\cdot|$ norm. Since the set of finite sequences are dense in $\ell^1$ with respect to the $|\cdot|_1$ norm, which dominates $|\cdot|$, you have that finite sequences are also dense in $X$.
Given a sequence $x$, write $P_Nx$ to be the finite sequence that equals $x$ in the first $N$ slots and $0$ afterward. That $x\in X$ if and only if $P_Nx \in X$ for all $N$ and $|P_N x|$ is uniformly bounded.
However, if $x\in \ell^2 \setminus \ell^p$ for any $p > 1$, then necessarily $|P_N x|_q$ for all $q \in [1,p]$ is not uniformly bounded, this shows that $x\in X$ implies that $x\in \ell^p$ for all $p > 1$.
And so Jochen's formulation holds, that
- $x\in X$ if and only if $x\in \cap_{1 < p \leq 2} \ell^p$ and $|x| < \infty$.
The general fact on the monotonicity of the $\ell^p$ norms also shows that
$$ (p-1)|x|_p = \int_1^p |x|_p ~dq \leq \int_1^p |x|_q ~dq \leq |x| $$
so that $X$ embeds into every $\ell^p$ continuously. Hence $X^*$ contains all $\ell^q$ for $q < \infty$. On the other hand, the embedding of $\ell^1$ into $X$ shows that $X^*$ is contained in $\ell^\infty$.
We can show that $X^*$ is strictly larger than $\cup_{q < \infty} \ell^q$:
- Let $y_k \in \ell^{2^k + 1}\setminus \ell^{2^{k-1} + 1}$ for $k \in \mathbb{N}_0$, be such that $|y_k|_{2^k + 1} \leq 2^{-k}$. Then $\sum_k y_k$ converges to some sequence in $\ell^\infty$.
- The pairing $|\langle x, \sum y_k\rangle| \leq \sum |x|_{1 + 2^{-k}} |y_k|_{1 + 2^k} \leq \sum 2^{-k} |x|_{1 + 2^{-k}} \approx |x|$ and so we see that $\sum y_k \in X^*$.
(I am hopeful that this example actually leads to a characterization of $X^*$, but am not sure how to prove it, nor what would be a nicer norm for it.)