The answer to the question exactly as you asked it is yes; your space is isomorphic as a vector space, with no topology, to various Banach spaces. (See various comments for details.)
Edit: The assertion that the answer is yes has met with vigorous disbelief. Also there's a technical point that I realized after some thought I simply didn't know how to do. I've added an Appendix at the bottom of this answer explaining things in detail, in particular showing that $\Bbb R^{\Bbb N}$ and $\ell^2(\Bbb N)$ do in fact have algebraic dimension $c$. Back to the short version:
But that norm simply has nothing to do with the structure of the space as a space of sequences. There's no complete norm on the space $X$ of all real sequences that has the sort of properties you must have in mind. In particular: For $n\in\Bbb N$ define $\Lambda_n:X\to\Bbb R$ by $$\Lambda_n x=x_n.$$There is no complete norm on $X$ such that every $\Lambda_n$ is bounded. (That is, there is no complete norm such that for every $n$, $x_n$ depends continuously on $x$.) In fact there's no such norm, complete or not.
Suppose on the other hand that $||\Lambda_n||=c_n$. Define $$S_N=\sum_{n=1}^N 2^{-n}c_n^{-1}\Lambda_n.$$Then $||S_N||<1$ for every $N$, which says that for every $x$ we have $$\left|\sum_{n=1}^N2^{-n}c_n^{-1}x_n\right|\le||x||$$for every $N$. This is impossible; for any sequence $c_n$ there exists a sequence $x$ such that $$\sup_N\left|\sum_{n=1}^N2^{-n}c_n^{-1}x_n\right|=\infty.$$
Appendix: Two things: why those bases show that the answer to the question is yes, and why those bases exist. Note that here "basis" means basis in the pure linear algebra sense, often called "Hamel basis": If $B$ is a basis for $X$ then every $x\in X$ is equal to a unique linear combination of finitely many elements of $B$.
$\newcommand\S{\Bbb R^{\Bbb N}}$ $\newcommand\L{\ell^2(\Bbb N)}$
Assume for now that $\S$ and $\L$ both have bases of cardinality $c$. Then there is a bijection betweenn the two bases. This gives us an isomorphism $I:\S\to\L$, defined by mapping linear combinations of the elements of the basis for $\S$ to the corresponding linear combinations of the elements of the basis for $\L$. We can thus define a norm on $\S$ by $||x||_{\S}=||Ix||_{\L}$, and it is straightforward to verify that the new norm on $\S$ is complete (given a Cauchy sequence in $\S$, by definition the corresponding sequence in $\L$ is Cauchy, hence convergent, so that by definition the original sequence in $\S$ is convergent.)
Ok, I left out some details there. Every detail is trivial - this equals that be definition since this or that is an isomorphism.
How do we know there are bases of cardinality $c$? Both spaces have cardinality $c$ so a basis can be no larger than $c$; we need $c$ independent vectors. It seems clear that we somehow should get this from the fact that there are $c$ subsets of $\Bbb N$, but I didn't see how to get the independence. I asked the Banach-space guy at the office:
Main Lemma There exists a map $S:\Bbb R\to\mathcal P(\Bbb N)$ such that $S(r)$ is infinite for every $r$, and such that given finitely many distinct reals $r_1,\dots,r_n$ there exists $N$ so that the sets $S(r_1)\cap[N,\infty),\dots,S(r_n)\cap[N,\infty)$ are pairwise disjoint.
Proof Let $\Bbb Q=\{q_1,\dots\}$. For each $r\in\Bbb R$ choose a sequence $(q_{n_j})$ of distinct rationals with $q_{n_j}\to r$, and let $S(r)=\{n_j\}$. QED.
Now it's clear that $\{\chi_{S(r)}:r\in\Bbb R\}$ is a linearly independent subset of $\S$. And if $(e_1,\dots)$ is orthonormal in $\L$ then $\left\{\sum_{j\in S(r)}\frac{e_j}{j}:r\in R\right\}$ is a linearly independent subset of $\L$.
Best Answer
There are two interpretations of the question possible:
I will show, with references, that the answer to 1 is no, even if we do not require that $(E, \|\cdot\|)$ be complete, and the answer to 2 is yes, using the axiom of choice in the form of "every vector space has a basis".
Answer to number 1:
The first thing to observe is that the topology defined by $d$ is the same as the product topology on $S$ (equivalently, the topology of pointwise convergence). Continuity of the identity map from $(S,d)$ to $\mathbb{R}^{\mathbb{N}}$ is most easily proved by showing that the projection mappings $\pi_n : S \rightarrow \mathbb{R}$ defined by $\pi_n((x_i)) = x_n$ are continuous. Proving that the identity mapping is continuous in the other direction is most easily done using the observation that $0 \leq \frac{|x|}{1 + |x|} < 1$ for all real $x$, which means $d(0,(x_i))$ can be bounded by bounding only finitely many coordinates.
Then we need the fact that every closed bounded subset of $S$ is compact. Let's recall the definitions. Recall that in a topological vector space $E$, a 0-neighbourhood $N \subseteq E$ is a set such that there exists an open set $U$ such that $0 \in U \subseteq N$. A set $B \subseteq E$ is bounded if for all 0-neighbourhoods $N$ there exists a real number $\alpha > 0$ such that $B \subseteq \alpha N$. If $f : E \rightarrow F$ is a continuous linear map between topological vector spaces, and $B \subseteq E$ is bounded, then $f(B)$ is bounded (see e.g. Schaefer's Topological Vector Spaces I.5.4, or prove it directly using the definition of continuity of linear maps in terms of 0-neighbourhoods). So if $B \subseteq S$ is bounded, for all $n \in \mathbb{N}$, $\pi_n(S) \subseteq \mathbb{R}$ is bounded. Let $B$ be a closed, bounded set. Since $[-1,1]$ is a 0-neighbourhood in $\mathbb{R}$, for each $n \in \mathbb{N}$, the set $\pi_n(B) \subseteq [-\alpha_n,\alpha_n]$ for some real $\alpha_n > 0$. It follows that $B \subseteq \prod_{i=1}^\infty [-\alpha_n,\alpha_n]$, which is compact in $S$ by Tychonoff's theorem (because $(S,d)$ has the product topology). As $B$ is closed, it is also compact.
Now, in a normed space $(E, \|\cdot\|)$, the closed unit ball $U$ is bounded (every 0-neighbourhood $N$ contains an open ball of radius $\epsilon > 0$, so $U \subseteq (\epsilon^{-1} + 1)N$). By Riesz's lemma, if $E$ is infinite-dimensional, then $U$ contains a sequence for which the distance between the elements is bounded below by some number, which therefore has no Cauchy subsequence, and therefore no convergent subsequence (even in an incomplete metric space, every convergent sequence is Cauchy). In a compact metric space, every sequence has a convergent subsequence, so this proves $U$ is not compact. Therefore there is no norm defining the topology of $(S,d)$.
Answer to number 2:
Since we are disregarding the original topology of $S$ in this section, we will be using the non-topological notions of basis and dimension. If $E$ is a vector space over $\mathbb{R}$, we say that a family $(x_i)_{i \in I}$ is a (Hamel) basis if:
The key facts are that every linearly independent set can be extended to a basis, all bases of a vector space $E$ have the same cardinality (called the dimension of $E$), and vector spaces are linearly isomorphic iff they have the same dimension. Unfortunately, standard references on linear algebra usually only prove these facts for the finite-dimensional case, but the general case is treated in Chapter IX of volume II of Jacobson's Lectures in Abstract Algebra.
It is clear that the dimension of $E$ is $\leq |E|$, because a linearly independent set in $E$ is a subset of $E$. We have $$ |S| = |\mathbb{R}|^{|\mathbb{N}|} = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0 \times \aleph_0} = 2^{\aleph_0}, $$ so the dimension of $S$ is $\leq 2^{\aleph_0}$. For any separable Banach space $E$, such as $\ell^2$, every element is the limit of a sequence from a countable dense subset, so $\mathbb{N}^{\mathbb{N}}$ maps surjectively onto $E$, so $E$ also has cardinality and therefore dimension $\leq 2^{\aleph_0}$.
In his paper On Infinite-dimensional Linear Spaces, Mackey proves in Theorem I.1 that every infinite-dimensional Banach space is of dimension $\geq 2^{\aleph_0}$. He does this in two steps. The first is to show that for every infinite-dimensional Banach space $E$, there exists an injective linear map $f : \ell^\infty \rightarrow E$. The second is the observation that the $(0,1)$-indexed family $((\alpha^i)_{i \in \mathbb{N}})_{\alpha \in (0,1)}$ in $\ell^\infty$ is linearly independent, making the dimension of $E$ greater than or equal to $|(0,1)| = 2^{\aleph_0}$. Therefore any separable infinite-dimensional Banach space has dimension exactly $2^{\aleph_0}$. Since $\ell^\infty$ is a linear subspace of $S$, this also proves that $S$ has dimension exactly $2^{\aleph_0}$.
So by mapping a basis of $S$ onto a basis of $\ell^2$ and extending by linearity, we can define a linear isomorphism $f : S \rightarrow \ell^2$. Let's use $\| \cdot \|_2$ for the norm on $\ell^2$ and define, for $x \in S$, $\|x\|_1 = \|f(x)\|_2$. The linearity of $f$ ensures that $\|\cdot\|_1$ is a norm. If $(x_i)_{i \in \mathbb{N}}$ is $\|\cdot\|_1$-Cauchy, define $y_i = f(x_i)$, and expanding the definitions shows that $(y_i)_{i \in \mathbb{N}}$ is $\|\cdot\|_2$-Cauchy, so converges to some $y \in \ell^2$. Since $f$ is an isomorphism, there exists $x \in S$ such that $f(x) = y$, and expanding the definitions and using linearity of $f$ shows that $(x_i)_{i \in \mathbb{N}}$ converges to $x$ in $\|\cdot\|_1$. Therefore $(S,\|\cdot\|_1)$ is a Banach space.
Some examples in mathematics are mainly used to show why things are defined the way they are, rather than for practical use. This is one of those examples.