How to draw a higher-genus surface

Quick and dirty: look at the boundary of a tubular neighborhood of a union of circles.

circle[x_, n_: 32] := {x + Cos[#], Sin[#], 0} & /@ Range[0, 2 [Pi], 2 [Pi]/n];
Graphics3D[Tube[circle[#, 72], .5] & /@ Range[-3, 3, 2], Boxed -> False]

Space them approximately two units apart (using x) and keep their radii less than $1/2$.

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For smooth surfaces--albeit at a price--we may subvert RegionPlot3D to do our work. It's a similar idea, only now we apply a 3D buffer to a circular skeleton rather than using tubular neighborhoods of fixed radius:

d[{x_, y_, z_}, x0_: 0] := Block[{u, v}, {u, v} = {x0, 0} + Normalize[{x - x0, y}]; 
  Norm[{u, v, 0} - {x, y, z}]^2];
RegionPlot3D[Min[d[{x, y, z}, #] & /@ Range[-2, 2, 2]] <= 1/2, {x, -4,4}, {y, -2,2}, {z, -2,2}, 
  BoxRatios -> {4, 2, 2}, Mesh -> None, PlotPoints -> 50, Boxed -> False, Axes -> False]

The argument x0 to d shifts the skeleton's center to x0 along the x-axis. Taking a contour of the shortest distance to a collection of circular skeletons does the job.

Here's a double torus by Stan Wagon from the 2010 one-liner competition

ContourPlot3D[(x^4 - x^2 + y^2)^2 + 9 z^2 == 0.04,
              {x, -1.2, 1.2}, {y, -1, 1}, {z, -0.4, 0.4}]

With Boxed, Axes, and Mesh set to False and the equation = 0.03:

The following pokes n holes in flattish blob:

genus[n_] := Module[{pts, fn},
  pts = If[n == 1, {0, 0}, 
    Table[2 {Cos[t], Sin[t]}, {t, 2 [Pi]/n, 2 [Pi], 2 [Pi]/n}]];
  fn = 10 z^2 + 
    Total[Join[#/n, (2 + 2/n)/#] &[#.# &[{x, y} - #] & /@ pts]]; 
  ContourPlot3D[fn == 18, {x, -4, 4}, {y, -4, 4}, {z, -2.5, 2.5}, 
   Mesh -> None, ContourStyle -> Yellow, BoxRatios -> Automatic, 
   Boxed -> False, Axes -> False]
  ]

Note: The expression fn is $10,z^2$ plus the sum over all points pts of $k,d^2 + l/d^2$, where $d$ is the distance to the point (dropping $z$ coordinates) and $k$, $l$ are coefficients depending on the number of holes $n$. The upshot is that the function goes to infinity at the vertical lines through the points and as $(x,y,z)$ moves away from the points.

With[{n = 1}, 
 10 z^2 + Total[Join[#/n, (2 + 2/n)/#] &[#.# &[{x, y} - #] & /@ {{a, b}}]]]

(* -> (-a + x)^2 + (-b + y)^2 + 4/((-a + x)^2 + (-b + y)^2) + 10 z^2 *)

Wagon's formulation of the double torus, as featured in David's answer, is based on the Gerono lemniscate. I find that using Booth's lemniscate (of which Bernoulli's is a special case) instead, with its adjustable parameters, to be more flexible. Here are two examples:

With[{a = 5, b = 4, c = 5, f = 3/4}, 
     ContourPlot3D[((x^2 + y^2)^2 - a x^2 + b y^2)^2 + c z^2 == f,
                   {x, -5/2, 5/2}, {y, -1, 1}, {z, -1/2, 1/2}, 
                   BoxRatios -> Automatic]]

With[{a = 3, b = 3, c = 10, f = 1/2}, 
     ContourPlot3D[((x^2 + y^2)^2 - a x^2 + b y^2)^2 + c z^2 == f,
                   {x, -2, 2}, {y, -1, 1}, {z, -1/2, 1/2},
                   BoxRatios -> Automatic]]

Of course, one can put all this in a Manipulate[] to explore the effects of changing the parameters, but I'll leave that up to the sufficiently interested reader.

Similar things can be done with e.g. rose curves or sinusoidal spirals if higher genus surfaces are desired.

By using the Erdős lemniscate of order n:

erdos[z_, n_] := Abs[z^n - 1]^2 - 1;
f[x_, y_, z_] := 
  erdos[x + I y, 3]^2 + (16*Abs[x + I y]^4 + 1)*(z^2 - 0.12^2);
ContourPlot3D[
 f[x, y, z] == 0, {x, -1.5, 1.5}, {y, -1.5, 1.5}, {z, -1.5, 1.5}, 
 ContourStyle -> {StippleShading[0.5], White}, Lighting -> "Accent", 
 PerformanceGoal -> "Quality", BoxRatios -> Automatic, Axes -> False, 
 Mesh -> None, Boxed -> False]