For your apparent purpose, in dimension $n$ it is convenient to begin with a homogeneous polynomial of total degree $2n$ with all individual exponents even. Then, for translates and rotations, all sorts of lower degree and odd exponent terms may show up.
In $\mathbb R^2,$ a rounded version of an ordinary square is
$$ A(x^4 + y^4) + B x^2 y^2 = 1.$$ The ordinary unit circle is $A=1, B = 2.$ Disjoint hyperbolas are $A=1, B=-2.$ A somewhat squared shape, indeed the $L^4$ "unit circle," is $A=1, B=0.$ An alternative to Piet Hein's "superellipse" is
$A=1, B=1.$ Finally, a real analytic curve that passes through all 8 lattice points with
$ |x| \leq 1,\; |y| \leq 1$ other than the origin itself is $A=1, B = -1,$ or
$$ x^4 - x^2 y^2 + y^4 = 1.$$
At some point I wanted a smooth version of an ordinary cube in $\mathbb R^3,$ meaning that it passed through all
26 integer lattice points with
$ |x| \leq 1,\; |y| \leq 1,\; |z| \leq 1$ other than the origin itself.
I wrote
$$ A( x^6 + y^6 + z^6) + B (y^4 z^2 + z^4 x^2 + x^4 y^2 + y^2 z^4 + z^2 x^4 + x^2 y^4) + C x^2 y^2 z^2 = 1.$$
To find $A,B,C$ it is only necessary to check the $(x,y,z)$ triples $(0,0,1),(0,1,1),(1,1,1),$ and evidently
$A=1, B=-\frac{1}{2}, C=1$ works. so
$$ ( x^6 + y^6 + z^6) - \frac{1}{2}(y^4 z^2 + z^4 x^2 + x^4 y^2 + y^2 z^4 + z^2 x^4 + x^2 y^4) + x^2 y^2 z^2 = 1$$ is a rounded cube. I recall graphing this in spherical coordinates with $\rho$ a function of $\theta, \phi.$ The trouble was that it is very flat near the axes, so without spherical coordinates many different patches were necessary. It is obvious that this is a star-shaped body around the origin, it needs just a little more work to confirm that it is compact.
