Remark 4.5.3 in Kalton and Albiac's Topics in Banach Space Theory states:
" ... since $C(K)^*$ is isometric to $\ell_1$ for every countable compact metric space $K$, the Banach space $\ell_1$ is isometric to the dual of many nonisomorphic Banach spaces."
With regard to the above, from H. P. Rosenthal's article in The Handbook of the Geometry of Banach Spaces, Vol 2, a result of Bessaga and Pełczyński is given:
Let $K$ be an infinite countable compact metric space.
$\ \ \ $(a) $C(K)$ is isomorphic to $C(\omega^{\omega^\alpha}+)$ for some countable ordinal $\alpha\ge0$.
$\ \ \ $(b) If $0\le\alpha<\beta<\omega_1$, then $C(\omega^{\omega^\alpha}+)$ is not isomorphic to $C(\omega^{\omega^\beta}+)$
See also this post at MathOverflow which gives an example of a separable Banach space $X$ and a non-separable Banach space $Y$ such that $X^*$ and $Y^*$ are isometrically isomorphic.
Preduals of $\ell_1$ are very interesting creatures. The question is slightly ill-posed but let me comment on that anyway.
We have quite a lot of isometric preduals of $\ell_1$. Indeed, for any countably infinite ordinal number $\alpha$ (endowed with the order topology) we have $C_0(\alpha)^* \cong \ell_1$. This essentially follows from the Riesz–Markov–Kakutani representation theorem as every measure on such a space has to be atomic.
Note that $c_0 = C_0(\omega)$ and $c=C(\omega+1)$ are isomorphic. Actually for each $\alpha\in [\omega, \omega^\omega)$ we have $C_0(\alpha) \cong c_0$. However, there exist $\aleph_1$ many countable ordinals which give pair-wise non-isomorphic Banach spaces. To be more precise, if $\alpha, \beta$ are countable ordinals then $C(\omega^{\omega^\alpha}+1)$ and $C(\omega^{\omega^\beta}+1)$ are isomorphic if and only if $\alpha = \beta$.
Historically, the first example different from the above-mentioned ones was due to Y. Benyamini and J. Lindenstrauss:
Y. Benyamini and J. Lindenstrauss. A predual of $\ell_1$ which is not isomorphic to a $C(K)$ space, Israel J. Math. 13 (1972), 246–254.
Johnson and Zippin proved that every isometric predual is a quotient of $C(\Delta)$, where $\Delta$ is the Cantor set.
W.B. Johnson and M. Zippin, Every separable predual of an $L_1$-space is a quotient of $C(\Delta)$, Israel J. Math. 16 (1973), 198–202.
This result combined with an old result of Pełczyński
A. Pełczyński, Banach spaces on which every unconditionally converging operator is weakly compact, Bull. Acad. Pol. Sci. Ser. Math. Astr. Phys. 10 (1962), 265–270.
asserting that an operator $T\colon C(K)\to X$ is weakly compact if and only if it is not bounded below on any isomorphic copy of $c_0$ yields the following corollary:
Corollary. Each isometric predual of $\ell_1$ contains a subspace isomorphic to $c_0$.
In some sense this answers OP's question.
One may wonder whether the Cantor set in the statement of the Johnson–Zippin theorem may be replaced by a countable compact Hausdorff space. This is not the case as shown by D. Alspach:
D. E. Alspach, A $\ell_1$-predual which is not isometric to a quotient of $C(\alpha)$, Contemp. Math. 144, Amer. Math. Soc., (1993), 9–14.
On the other hand, Gasparis constructed some new preduals of $\ell_1$ which are quotients of $C(\alpha)$:
I. Gasparis, A class of $\ell_1$-preduals which are isomorphic to quotients of $C(\omega^\omega)$, Studia Math. 133 (2) (1999), 131–143.
Since there is no characterisation of complemented subspaces of $C(K)$-spaces, this result is also noteworthy:
I. Gasparis, A new isomorphic $\ell_1$ predual not isomorphic to a complemented subspace of a $C(K)$ space, Bull. London Math. Soc. 45 (2013), 789–799.
If you are interested in spaces whose dual space is only isomorphic to $\ell_1$ then the situation is even more exciting. J. Bourgain and F. Delbaen constructed isomorphic preduals without subspaces isomorphic to $c_0$:
J. Bourgain and F. Delbaen, A class of special $\mathscr{L}_\infty$-spaces, Acta Math. 145 (1980), 155–176.
A recent variation of the BD construction is the famous Argyros–Haydon space which is an isomorphic $\ell_1$-predual with the property that each operator on this space is of the form $cI + K$, where $c$ is a scalar and $K$ is a compact operator.
S.A. Argyros and R.G. Haydon, A Hereditarily Indecomposable $\mathscr{L}_\infty$-space that solves the scalar-plus-compact problem, Acta Math. 206 (2011), no. 1, 1–54.
Even more recently, Spiros A. Argyros, Ioannis Gasparis and Pavlos Motakis constructed a $c_0$-asymptotic $\ell_1$-predual without copies of $c_0$:
S.A. Argyros, I. Gasparis and P. Motakis, On the structure of separable $\mathcal{L}_\infty$-spaces, Mathematika 62 (2016), no. 3, 685–700.
A nice overview of the BD-spaces (and some new variants of the Argyros–Haydon space) can be found in Matt Tarbard's PhD thesis:
M. Tarbard, Operators on Banach Spaces of Bourgain–Delbaen Type, PhD Thesis, University of Oxford, 2012.
I would also recommend the following paper by M. Daws, R. Haydon, Th. Schlumprecht and S. White
M. Daws, R. Haydon, Th. Schlumprecht, and S. White, Shift invariant preduals of $\ell_1(\mathbb{Z})$, Israel Journal of Mathematics, 192 (2012), 541–585.
which explains these subtleties very well.
Laustsen and myself have exhibited a version of the Argyros–Haydon space whose algebra of bounded operators has certain peculiar properties:
T. Kania and N.J. Laustsen, Ideal structure of the algebra of bounded operators acting on a Banach space, Indiana University Mathematics Journal 66 (2017), 1019–1043.
Let me finish with an open problem (I think) which I like very much:
Problem. We can easily construct $\aleph_1$ many isometric preduals of $\ell_1$. Can we construct in $\mathsf{ZFC}$ continuum many isometric preduals of $\ell_1$?
Best Answer
$c_0$ is not the dual of some normed vector space.
sketch of proof:
This can be proven using the Krein-Milman theorem.
If $c_0$ was the dual of a normed vector space, then its unit ball would be weakly-* compact.
However, it can be shown that the unit ball of $c_0$ has no extremal points, therefore by Krein-Milman it is not weakly-* compact.
edit: for $c$ the situation is more complicated than I initially thought.
As uniquesolution pointed out in the comments, its unit ball has many extreme points, and it is not clear to me if Krein-Milman can be used to show that the unit ball of $c$ cannot be weakly-* compact.