Let $f$ be the function defined by $f(x) = -2 \sin(\pi x)$.  How many values of $x$ such that $-2 \le x \le 2$ satisfy the equation $f(f(f(x))) = f(x)$?
The graph of $y = f(x)$ is shown below.

[asy]
unitsize(1.5 cm);

real func (real x) {
  return (-2*sin(pi*x));
}

draw(graph(func,-2,2),red);
draw((-2.5,0)--(2.5,0));
draw((0,-2.5)--(0,2.5));

draw((1,-0.1)--(1,0.1));
draw((2,-0.1)--(2,0.1));
draw((-1,-0.1)--(-1,0.1));
draw((-2,-0.1)--(-2,0.1));
draw((-0.1,1)--(0.1,1));
draw((-0.1,2)--(0.1,2));
draw((-0.1,-1)--(0.1,-1));
draw((-0.1,-2)--(0.1,-2));

label("$1$", (1,-0.1), S, UnFill);
label("$2$", (2,-0.1), S, UnFill);
label("$-1$", (-1,-0.1), S, UnFill);
label("$-2$", (-2,-0.1), S, UnFill);
label("$1$", (-0.1,1), W, UnFill);
label("$2$", (-0.1,2), W, UnFill);
label("$-1$", (-0.1,-1), W, UnFill);
label("$-2$", (-0.1,-2), W, UnFill);

label("$y = f(x)$", (2.8,1), red);
[/asy]

The equation $f(x) = 0$ has five solutions in $[-2,2].$  For a fixed nonzero real number $y,$ where $-2 < y < 2,$ the equation $f(x) = y$ has four solutions in $[-2,2].$

We want to solve the equation
\[f(f(f(x))) = f(x).\]Let $a = f(x),$ so
\[a = f(f(a)).\]Let $b = f(a),$ so $a = f(b).$  Thus, both $(a,b)$ and $(b,a)$ lie on the graph of $y = f(x).$  In other words, $(a,b)$ lie on the graph of $y = f(x)$ and $x = f(y).$

[asy]
unitsize(1.5 cm);

real func (real x) {
  return (-2*sin(pi*x));
}

draw(graph(func,-2,2),red);
draw(reflect((0,0),(1,1))*(graph(func,-2,2)),blue);
draw((-2.5,0)--(2.5,0));
draw((0,-2.5)--(0,2.5));

draw((1,-0.1)--(1,0.1));
draw((2,-0.1)--(2,0.1));
draw((-1,-0.1)--(-1,0.1));
draw((-2,-0.1)--(-2,0.1));
draw((-0.1,1)--(0.1,1));
draw((-0.1,2)--(0.1,2));
draw((-0.1,-1)--(0.1,-1));
draw((-0.1,-2)--(0.1,-2));

label("$y = f(x)$", (2.8,0.6), red);
label("$x = f(y)$", (2.8,-0.5), blue);
[/asy]

Apart from the origin, there are 14 points of intersection, all of which have different $x$-coordinates, strictly between $-2$ and 2.  So if we set $(a,b)$ to be one of these points of intersection, then $a = f(b)$ and $b = f(a).$  Also, the equation $f(x) = a$ will have four solutions.

For the origin, $a = b = 0.$  The equation $f(x) = 0$ has five solutions.

Therefore, the equation $f(f(f(x))) = f(x)$ has a total of $14 \cdot 4 + 5 = \boxed{61}$ solutions.