The closed unit ball of $c_0$ has no extreme points. The closed unit ball of $c$ has many extreme points, such as $(1,1,\ldots)$. Since the property of being an extreme point is preserved by isometries, $c$ and $c_0$ are not isometrically isomorphic.
For simplicity lets assume $\|\hat f\|_1 = 1$.
Let $F_0 \in (l^\infty)^*$ be defined as $F_0(x) = \sum \hat f_k x_k$ and $F$ be any extension of $f$ that disagrees with $F_0$.
Let $x = (x_1, x_2, \ldots) \in l^\infty$ have norm of $1$ be a vector on which $F_0$ and $F$ disagree: $F(x) - F_0(x) > \varepsilon > 0$ (if difference is less then $0$ - replace $x$ with $-x$).
Let $N$ be such that $\sum\limits_{k=1}^N |\hat f_k| > 1 - \frac{\varepsilon}{2}$.
Let $y = (0, 0, \ldots, 0, x_{N + 1}, \ldots)$. We have $\|y\| \leqslant 1$, $F(y) - F_0(y) > \varepsilon$ and $|F_0(y)| < \frac{\varepsilon}{2}$, so $F(y) > \frac\varepsilon 2$.
Let $z = (\operatorname{sgn}\hat f_1, \operatorname{sgn}\hat f_2, \ldots, \operatorname{sgn}\hat f_{N}, 0, 0, \ldots)$. As $F$ is extension of $f$, we have $F(z) = \sum\limits_{k=1}^N |\hat f_k| > 1 - \frac{\varepsilon}{2}$
Then we have $\|y + z\| = 1$. At the other hand, we have $F(y + z) = F(y) + F(z) > \frac{\varepsilon}{2} + 1 - \frac{\varepsilon}{2} = 1$. This implies $\|F\| > 1$.
So any extension that disagrees with $F_0$ has norm greater then $1$ - so $F_0$ is the only extension of $f$ that has the same norm.
Best Answer
Consider the mapping $l\colon c\to\mathbb{R}$, $l((x_n)):=\lim_{n\to\infty} x_n$. This mapping is linear and surjective. The kernel equals $c_0$. Thus $c/c_0\sim l(c)=\mathbb{R}$.