Is every Banach space densely embedded in a Hilbert space? Can every Banach space be densely embedded in a Hilbert space? This is clear if the Banach space is actually a Hilbert space, but much can you relax this?
If the embedding exists, is the target Hilbert space unique?

We can show that $\ell_\infty(I)$ cannot be embedded into a Hilbert space, for an uncountable index set $I$.
Here, I am interpreting the question as Normal Human suggests - where the embedding is assumed to be linear and continuous. Suppose that $f\colon \ell_\infty(I)\to H$ is an embedding into Hilbert space $H$. By continuity, there exists a $K\in\mathbb{R}$ such that $\lVert f(x)\rVert\le K\lVert x\rVert$ for all $x\in\ell_\infty(I)$.
Given any sequence $x_1,\ldots,x_n\in H$, the identity
$$
\sum_{\epsilon_1,\ldots,\epsilon_n=\pm1}\left\lVert\sum_{r=1}^n\epsilon_rx_r\right\rVert^2=2^n\left(\sum_{r=1}^n\lVert x_r\rVert^2\right)
$$
holds. This implies that there exists a sequence $\epsilon_r\in\{\pm1\}$ such that
$$
\left\lVert\sum_{r=1}^n\epsilon_rx_r\right\rVert^2\ge\sum_{r=1}^n\lVert x_r\rVert^2.
$$
Now, for each $i\in I$, let $e_i\in\ell_\infty(I)$ be defined by $(e_i)_j=0$ for $j\not=i$ and $(e_i)_i=1$. Also, for each $n\in\mathbb{Z}_{>0}$, let $S_n$ be the set of $i\in I$ such that $\lVert f(e_i)\rVert\ge1/n$. As $f$ is an embedding, we have $f(e_i)\not=0$, so $\bigcup_{n=1}^\infty S_n=I$ is uncountable. Hence, $S_n$ is infinite for some $n$. Then, for any $N > 0$, pick a sequence $i_1,\ldots,i_N$ of distinct elements of $S_n$. By what we showed above, there is a sequence $\epsilon_1,\ldots,\epsilon_N\in\{\pm1\}$ such that $x\equiv\sum_{r=1}^N\epsilon_re_{i_r}$ satisfies
$$
\lVert f(x)\rVert^2\ge\sum_{r=1}^N\lVert f(e_{i_r})\rVert^2\ge N/n^2.
$$
However, $\Vert x\rVert=1$, so
$$
\lVert f(x)\rVert\ge n^{-1}\sqrt{N}\lVert x\rVert.
$$
Choosing $N>K^2n^2$ gives a contradiction.
