Well, about time I rewrote this answer to address all the issues raised the comments. It gets pretty long though, because there are too many subtleties. So here's the result in advance: the optimal curve is composed of circular arcs between pairs of points, with all arcs having the same radius. Unless such a construction leads to self-intersections, in which case all bets are off.
In general, we may consider a sequence of $n$ points $P_0, P_1, P_2, \ldots, P_n=P_0$ on the plane. We want to find the closed curve of specified length $L$ that connects these points in the specified order and maximizes the area $A$ enclosed by the curve.
For area to be well-defined, of course, the curve has to be simple (i.e. without self-intersections) or "weakly simple" (i.e. it can be made simple by an infinitesimal perturbation). I believe that it is possible, in some configurations of points, that the optimal curve is only weakly simple.¹ However, as such cases involve "contact interactions" between multiple segments that may be topologically distant, they are much harder to analyze than cases where optimality is achieved by simple curves. In this answer, therefore, I will give up a little generality, and only characterize the optimal curve in cases where it is simple.
A simple closed curve $\mathcal C$ that connects the points $P_i$ in the specified order is composed of $n$ curves $\mathcal S_i$ from $P_i$ to $P_{i+1}$, each simple and mutually non-intersecting (apart from meeting at the endpoints). For each curve $\mathcal S_i$, let $\ell_i$ be its length, and $A_i$ be the signed area between it and the line segment $P_i P_{i+1}$. Then the entire closed curve $\mathcal C$ has length $\ell = \sum_{i=0}^{n-1}\ell_i$ and area $A = \operatorname{area}(P_0 P_1 \cdots P_{n-1}) + \sum_{i=0}^{n-1}A_i$. (I've assumed for simplicity that the polygon $P_0 P_1 \cdots P_{n-1}$ itself is weakly simple and so has a well-defined area, but I think this restriction could be lifted with a more careful treatment.) Since the area of the polygon does not change, maximizing $A$ amounts to maximizing $\sum_{i=0}^{n-1}A_i$.
So, the problem is to find simple curves $\mathcal S_i$ that maximize $\sum_i A_i$ under the constraints that $\sum_i\ell_i = L$ and the $\mathcal S_i$ are pairwise non-intersecting.
For now, let us lift the non-intersection constraint. If we find that in the relaxed problem, the maximum area is nevertheless attained by $\mathcal S_i$ that do not intersect, then that is certainly also the maximum area over all non-intersecting curves.² So now the curves $\mathcal S_i$ are basically independent, connected only by the constraint $\sum_i\ell_i = L$.
Lemma 1: Each $\mathcal S_i$ is a circular arc.
Hold the rest of the curve fixed and consider variations in $\mathcal S_i$. Its end points are $P_i$ and $P_{i+1}$, its length is fixed at $L-\sum_{j\ne i}\ell_j$, and we wish to maximize the area $A_i$. Such a curve must be a circular arc. ∎
Lemma 2: $\mathcal S_i$ and $\mathcal S_{i+1}$ have the same radius.
Without loss of generality, we may take $i=0$. Again, hold the rest of the curve fixed and consider variations in $\mathcal S_0$ and $\mathcal S_1$. We wish to maximize $A_0+A_1$ under the constraint that $\ell_0+\ell_1$ is fixed at $L'=L-\sum_{j\ge2}\ell_j$.
First, a simple geometrical argument that does not cover all cases but provides intuition. Observe that the relative orientation of $P_0P_1$ and $P_1P_2$ doesn't matter; if we have another point $P_2'$ with $|P_1P_2|=|P_1P_2'|$, any arc $\mathcal S_1$ from $P_1$ to $P_2$ can be rotated to an arc $\mathcal S_1'$ from $P_1$ to $P_2'$ with the same length and enclosed area. So arcs $\mathcal S_0$ and $\mathcal S_1$ are optimal for $P_0,P_1,P_2$ if and only if $\mathcal S_0$ and $\mathcal S_1'$ are optimal for $P_0,P_1,P_2'$. Now when $L'$ is sufficiently small,³ all feasible arcs $\mathcal S_0$ and $\mathcal S_1$ are less than semicircular, and there is a unique pair $(\mathcal S_0,\mathcal S_1)$ whose radii $r_0$ and $r_1$ are equal. Then we can pick $P_2'$ such that $\mathcal S_0$ and $\mathcal S_1'$ form a continuous arc of a single circle. By lemma 1, this curve encloses the largest area among all curves of length $L'$ from $P_0$ to $P_2'$; certainly then it encloses the largest area among such curves from $P_0$ through $P_1$ to $P_2'$. The result follows.
This argument does not work for large $L'$, in which case there may be more than one possible $(\mathcal S_0,\mathcal S_1)$ with $r_0=r_1$, and the combined arc $\mathcal S_0\mathcal S_1'$ may span more than a full circle. To cover all cases, we resort to analysis. A circular arc $\mathcal S$ between two fixed points a distance $d$ apart has exactly one degree of freedom, which can be parametrized by its length $\ell$, its radius $r$, or its central angle $\theta$. We have the relations $d = 2r\sin(\theta/2)$, $\ell = r\theta$, and $A = \frac12r^2(\theta-\sin\theta)$. Since $d$ is constant, this gives
$$\frac{\mathrm dr}{\mathrm d\theta} = -\frac12 r\cot\frac\theta2,$$
from which we can derive
$$\begin{align}
\frac{\mathrm d\ell}{\mathrm d\theta} &= r\left(1 - \frac\theta2\cot\frac\theta2\right), \\
\frac{\mathrm dA}{\mathrm d\theta} &= r^2\left(1 - \frac\theta2\cot\frac\theta2\right),
\end{align}$$
and so
$$\frac{\mathrm dA}{\mathrm d\ell} = r.$$
(It would be nice to see a geometrical proof of this result.) Coming to the case of $\mathcal S_0$ and $\mathcal S_1$, since $\ell_0 + \ell_1 = L'$, we have $\mathrm d\ell_0 + \mathrm d\ell_1 = 0$, while for area to be maximized we require $\mathrm dA_0 + \mathrm dA_1 = 0$. This implies that at the maximum, we must have $r_0 = r_1$. ∎
From the above lemmas, it follows that the closed curve $\mathcal C$ that maximizes the area sum $\sum_{i=1}^n A_i$ is composed of $n$ circular arcs of equal radius $r$. If this curve $\mathcal C$ is simple, then it is also the closed curve that encloses the largest area, which is what was desired.
In your example with $n=3$ and points $P_0=(-1,0)$, $P_1=(0,0)$ and $P_2=(1,0)$, I numerically found the solution to be $r\approx1.30889$ with arcs of central angle $\theta_0=\theta_1\approx44.9^\circ$ and $\theta_2\approx260.3^\circ$. This yields an area of about $4.87$ units, which is greater than that of your cardioid. Here is a plot comparing the numerically optimized solution in blue and the cardioid in red.
And here is how the optimal curves vary with $L$:
¹ For example, consider the Pac-Man-like shape $P_0=(-1+\epsilon,0)$, $P_i=\bigl(\cos(2\pi i/n),\sin(2\pi i/n)\bigr)$ for $i=1,\ldots,n-1$.
² This turns out to be the case often enough -- in fact, I would guess that it is always true for convex polygons. But on the other hand, if the optimal solution to the relaxed problem yields intersecting $\mathcal S_i$, then the result probably does not tell us anything about the solution to the original problem.
³ In particular, when $L' \le \min(d_0 + \frac\pi2 d_1, d_1 + \frac\pi2 d_0)$ where $d_0$ and $d_1$ are the lengths $|P_0P_1|$ and $|P_1P_2|$ respectively.
