Does there exist a 4D torus with a spherical cross-section, analogous to a circle for the 3D case? I don't mean to be a bother. 
It seems as though the answer may be obvious, but then, seemingly simple math questions can have surprising answers.
I should also like any pointers re: the general case for the torus. 
Thanks in advance for any (polite) suggestions.

 A: For any $d\geq1$ a $d$-dimensional torus $T^d$ is a $d$-dimensional manifold whose most basic model is ${\mathbb T}^d:={\mathbb R}^d/{\mathbb Z}^d$. Think of a $d$-dimensional cube whose  opposite $(d-1)$-dimensional faces are identified in pairs. The one-dimensional torus ${\mathbb T}={\mathbb R}/{\mathbb Z}$ happens to be the same thing as the one-sphere, or circle, $S^1$. The two-dimensional torus appears in many computer games as a rectangular board where creatures disappearing on the right surprisingly reappear on the left moving in the same direction.
A familiar realization of a $2$-torus $T^2$ is the surface
$$S:\quad \bigl(\sqrt{x^2+y^2}- a\bigr)^2 + z^2=b^2,\qquad a>b>0,$$
in ${\mathbb R}^3$. A parametric representation of $S$ is given by
$$(\phi,\theta)\mapsto\bigl((a+b\cos\theta)\cos\phi, \>(a+b\cos\theta)\sin\phi, \>b\sin\theta\bigr)\qquad\bigl(\phi\in{\mathbb R}/(2\pi{\mathbb Z}), \ \theta\in{\mathbb R}/(2\pi{\mathbb Z})\bigr)\ ,$$
which exhibits $S$ as bijective copy of the basic model ${\mathbb T}^2$ of such a surface.
The donut you can eat, i.e., the $3$d body
$$B:=\bigl\{(x,y,z)\>\bigm|\> \bigl(\sqrt{x^2+y^2}- a\bigr)^2 + z^2\leq b^2\bigr\},\qquad a>b>0,$$
is called a full torus.
The three-dimensional object you have in mind is not the three-dimensional torus $T^3\sim T^1\times T^1\times T^1$, but is the product of a $2$-sphere $S^2$ with $T^1$. You can envisage this object in the form of a spherical shell (with positive thickness) in ${\mathbb R}^3$ whose inner and outer spherical surfaces have been identified in the obvious way. Going around the $x_4$-axis in your hypothetical picture then corresponds to traversing the shell outwards in radial direction and automatically being teleported to the inner boundary when arriving at the outer boundary.
A: I'm way late to the party as usual, but I've got an equation for you, and some nice visuals of this object. Christian already gave the specifics a few ways. As you know, a torus can be constructed by starting with a circle in the xy plane:
$$x^2 + y^2 = r^2$$
Translate the circle by R units along the x-axis, which should be around $R = 2\cdot r$, for a ring torus,
$$(x-R)^2 + y^2 = r^2$$
Then, sweep the circle around a circular path into 3D, by replacing $x$ with $\sqrt{x^2+z^2}$ , 
$$\left(\sqrt{x^2+z^2}-R\right)^2 + y^2 = r^2$$
which gives us the equation of a torus.

To construct the $S^2$ x $S^1$ , we start with a sphere in plane $xyz$:
$$x^2 + y^2 + z^2 = r^2$$
Shift along $x$ by $R$ units where $R = 2\cdot r$,
$$(x-R)^2 + y^2 + z^2 = r^2$$
Sweep around in a circle, into 4D:
$$\left(\sqrt{x^2+w^2}-R\right)^2 + y^2 + z^2 = r^2$$

Cross Sections of $S^2$ x $S^1$
Taking some sections of this object, we can set either $w=0$ , or $z=0$:
$$\left(\sqrt{x^2}-R\right)^2 + y^2 + z^2 = r^2$$
This cross-section will give us the disjoint pair of spheres that you're looking for, centered at $(+R,0,0)$ and $(-R,0,0)$, since $\left(\sqrt{x^2}-R\right)^2$ has two values of $(x-R)^2$ and $(x+R)^2$

$$\left(\sqrt{x^2+w^2}-R\right)^2 + y^2 = r^2$$
This cross-section will give us a single torus. 

Now, what does this thing look like, as sliced in 3D? Using the rotate equation,
$$\left(\sqrt{x^2+\left(w\cdot \cos(t)+z\cdot \sin(t)\right)^2}-R\right)^2 + y^2 + \left(w\cdot \sin(t)-z\cdot \cos(t)\right)^2 = r^2$$
Animating the value of t from 0 to 2$\pi$:

Holding at different angles, and passing through a 3-plane:

