The space about whose double dual you are asking is usually denoted $\ell_1$.  It is a very well-understood space, and so is its dual $\ell_1^*=\ell_\infty$.  Many of the properties of $\ell_1^{**}=\ell_\infty^*$ are therefore inherited by virtue of it being the dual of a well-understood space.  For instance, it is well-known that every separable space embeds into $\ell_\infty$.

A nice introduction to $\ell_\infty^*$ is given in Diestel's book *Sequences and Series in Banach Spaces*, p76cc.  A very nice characterization is given there as follows.

Let $\Sigma$ be a $\sigma$-field of subsets of a set $\Omega$, and denote by $B(\Sigma)$ the set of bounded, $\Sigma$-measurable scalar-valued functions on $\Sigma$, and note that $B(\Sigma)$ is a Banach space when endowed with the sup norm $\|\cdot\|_\infty$.  Also denote by $ba(\Sigma)$ the set of all finitely additive scalar-valued signed measures with bounded total variation, and note that $ba(\Sigma)$ is also a Banach space, as long as it is endowed with the total variation norm $\|\cdot\|_1$.  We can now make the isometric identification $B(\Sigma)^*=ba(\Sigma)$ via the action $\mu(f)=\int f\,d\mu$ for all $\mu\in ba(\Sigma)$ and $f\in B(\Sigma)$.  If $\Omega=\mathbb{N}$ and $\Sigma$ is the counting measure then $B(\Sigma)=\ell_\infty$, and hence $\ell_\infty^*$ is precisely the space of finitely additive signed measures on $\mathbb{N}$ with bounded total variation, denoted $ba$ for short.  More information on this identification can be found in Dunford-Schwartz's *Linear Operators I* (on p296, it seems).

Note that, according to Diestel, you should read [this paper][1] to get a better understanding of the space $ba$.  However, I have not read it myself, so beware : )

There is another natural identification, due to the fact that $\ell_\infty=C(\beta\mathbb{N})$, where $\beta\mathbb{N}$ denotes the Stone-Cech compactification of $\mathbb{N}$.  Thus, $\ell_\infty^*$ can be identified with $C(\beta\mathbb{N})^*$, which in turn is identifiable with the space of regular Borel measures on $\beta\mathbb{N}$.  See chapter 15 of Carother's *A Short Course In Banach Space Theory* on this, especially remark 2 on p152.

Also, note that this question has been asked before on MO, [here][2].  If you're wondering about the title of the question, it's because $\ell_\infty\cong L_\infty[0,1]$, and so $\ell_\infty^*\cong L_\infty[0,1]^*$ (isomorphically, not isometrically).


  [1]: http://www.ams.org/journals/tran/1952-072-01/S0002-9947-1952-0045194-X/S0002-9947-1952-0045194-X.pdf
  [2]: https://math.stackexchange.com/questions/47395/the-duals-of-l-infty-and-l-infty