The accepted answer says, “this is impossible. PDF, and therefore TikZ, can only draw cubic bézier lines,” but I’m afraid that’s not quite true. If you convert to a coordinate system where the solution falls out simply and naturally, pgfplots can graph parameterized polar curves. (Technically, it does approximate them in terms of other primitives.)
A Bézier curve is mathematically equivalent to a cubic spline, and polar coordinates are just another coordinate system, so we can convert the control points of a Bézier curve to a piecewise cubic function in polar coordinates. And if you express the problem as a polar function, with the origin in the interior of your figure, adding ε to the radius of your function gives you a close approximation to the boundary of its ε-neighborhood. (See below for a proof.)
If we convert the control points you gave to the new coordinate system, that is, (θ, ρ) = (atan2(y′,x′), √(x′²+y′²)), where (x′, y′) are the translated Cartesian coordinates relative to the new origin we chose, we can interpolate ρ(θ) as a cubic-spline approximation. To make it a smooth, closed loop, we add 2π to the angle of the first two points, add these duplicates to the end of our list of points, and use natural boundary conditions.
Here is a plot of the original data points you gave (after translation), their polar cubic-spline interpolation, and its ε-neighborhood for ε=0.2.
\documentclass{standalone}
\usepackage{pgfplots}
\pgfplotsset{compat=1.17}
\usepgfplotslibrary{polar}
\begin{document}
\begin{tikzpicture}
\begin{axis}
\addplot [only marks] table {
-1.1 -0.6
-0.5 -0.8
0.3 -0.4
1.0773 -0.4579
0.9905 0.1074
0.4752 0.2828
-0.1 0.4
-0.9 0.0
};
\addplot [data cs=polarrad, domain=-2.6422459319096627:-2.129395642138459] (x, 1.2286380393395102*x^3+7.697565288852661*x^2+15.064811001675386*x+9.982031406751004);
\addplot [data cs=polarrad, domain=-2.129395642138459:-0.9272952180016125] (x, 0.5655789449594107*x^3+3.46181985069289*x^2+6.045233124450111*x+3.5799481315203536);
\addplot [data cs=polarrad, domain=-0.9272952180016125:-0.4019079929235029] (x, -2.8974117495348413*x^3-6.171824282272918*x^2+-2.8879990119783727*x+0.818700317717533);
\addplot [data cs=polarrad, domain=-0.4019079929235029:0.10800811798145088] (x, 1.3653634987432328*x^3-1.0320939493146357*x^2-0.8223003096910624*x+1.0954405908578588);
\addplot [data cs=polarrad, domain=0.10800811798145088:0.5368219472510897] (x, 1.1773690800521026*x^3-0.9711791792530982*x^2-0.8288795993626819*x+1.0956774630895545);
\addplot [data cs=polarrad, domain=0.5368219472510897:1.81577498992176] (x, -0.2685250430007833*x^3+1.3573839167153727*x^2-2.0789033748375023*x+1.319357528843072);
\addplot [data cs=polarrad, domain=1.81577498992176:3.14159265358979] (x, 0.1780847151433292*x^3-1.0754445705638838*x^2+2.33856574713336*x-1.354352454632371);
\addplot [data cs=polarrad, domain=3.141592653589793:3.6409393752699235] (x, -1.7982844692981617*x^3+17.55139616130403*x^2-56.17938025569*x^1+59.92549730057898);
\addplot [data cs=polarrad, domain=-2.6422459319096627:-2.129395642138459, gray] (x, 1.2286380393395102*x^3+7.697565288852661*x^2+15.064811001675386*x^1+9.982031406751004+0.2);
\addplot [data cs=polarrad, domain=-2.129395642138459:-0.9272952180016125, gray] (x, 0.5655789449594107*x^3+3.46181985069289*x^2+6.045233124450111*x+3.5799481315203536+0.2);
\addplot [data cs=polarrad, domain=-0.9272952180016125:-0.4019079929235029, gray] (x, -2.8974117495348413*x^3-6.171824282272918*x^2+-2.8879990119783727*x+0.818700317717533+0.2);
\addplot [data cs=polarrad, domain=-0.4019079929235029:0.10800811798145088, gray] (x, 1.3653634987432328*x^3-1.0320939493146357*x^2-0.8223003096910624*x+1.0954405908578588+0.2);
\addplot [data cs=polarrad, domain=0.10800811798145088:0.5368219472510897, gray] (x, 1.1773690800521026*x^3-0.9711791792530982*x^2-0.8288795993626819*x+1.0956774630895545+0.2);
\addplot [data cs=polarrad, domain=0.5368219472510897:1.81577498992176, gray] (x, -0.2685250430007833*x^3+1.3573839167153727*x^2-2.0789033748375023*x+1.319357528843072+0.2);
\addplot [data cs=polarrad, domain=1.81577498992176:3.14159265358979, gray] (x, 0.1780847151433292*x^3-1.0754445705638838*x^2+2.33856574713336*x-1.354352454632371+0.2);
\addplot [data cs=polarrad, domain=3.141592653589793:3.6409393752699235, gray] (x, -1.7982844692981617*x^3+17.55139616130403*x^2-56.17938025569*x^1+59.92549730057898+0.2);
\end{axis}
\end{tikzpicture}
\end{document}
A somewhat computationally-easier approximation would be to do what Jairo Bochi did and generate a new set of points at the boundary of the ε-neighborhood and draw a new loop through them. However, instead of estimating the delta between pairs of points, we can convert to polar, and add ε to the polar radius of each original point.
\documentclass{standalone}
\usepackage{pgfplots}
\pgfplotsset{compat=1.17}
\usepgfplotslibrary{polar}
\begin{document}
\begin{tikzpicture}
\begin{axis}
\addplot+ [fill, smooth cycle, mark=*, mark options={black}, draw=gray, fill=lightgray] coordinates {
(-1.2755791145828763, -0.6957704261361151)
(-0.605999788000636, -0.9695996608010178)
(0.4199999999999998, -0.56)
(1.261363262096108, -0.5361350020549593)
(1.1893345582858206, 0.12895964821796777)
(0.6470676596962922, 0.38508151128390444)
(-0.14850712500726604, 0.5940285000290663)
(-1.1000000000000003, 0)
};
\addplot+ [fill, smooth cycle, mark=*, mark options={black}, draw=black, fill=darkgray] coordinates {
(-1.1, -0.6)
(-0.5, -0.8)
(0.3, -0.4)
(1.0773, -0.4579)
(0.9905, 0.1074)
(0.4752, 0.2828)
(-0.1, 0.4)
(-0.9, 0.0)
};
\end{axis}
\end{tikzpicture}
\end{document}
Update
I received a comment claiming that this approach “dilates the function from the chosen origin, but that isn't the same as the epsilon neighbourhood.” This is of course an approximation of a cubic-spline interpolation that gives us a closed subset of the true neighborhood. All the other answers are approximations of interpolations, too. I think that’s fine: the purpose here is to visualize a diagram, not to find an exact analytical solution. But here is a proof that the approximation is mathematically justified.
For a circle, where r'(t) = 0 everywhere, the solution is exact. There might be others (the expression I took the limit of looks like you could multiply by a conjugate instead, for instance). I’m sure anything I came up with in a few minutes has been studied long, long before.
Of course, since our r(t) is a differentiable periodic piecewise function composed of cubic polynomials, we could calculate exact values of x(t), y(t) and their derivatives at every t.
I would like how to draw a 
