Solution 1. First note that $f(0) = c$, $f(1) = a + b + c$, $f(-1) = a - b + c$. Second, note that if $f(\alpha) = f(\beta)$, then $a(\alpha^2 - \beta^2) + b(\alpha - \beta) = 0$, so $(\alpha - \beta)(a(\alpha + \beta) + b) = 0$. Third, a key observation is that for fixed $k$, there are at most two distinct real numbers $x$ for which $f(x) = k$. It follows from this that since $f(f(1)) = f(f(0)) = f(f(-1))$, the set $\{f(1), f(0), f(-1)\}$ can contain at most two distinct numbers. This leads us to three cases.

Case 1: $f(1) = f(-1)$.
This gives $b = 0$, so $f(1) = f(-1) = a + c$ and $f(0) = c$. Therefore
$$0 = f(f(1)) - f(f(0)) = (f(1) - f(0))(a(f(1) + f(0)) + 0) = a^2(a + 2c).$$
So $c = -a/2$, which yields $f(x) = a(x^2 - \frac{1}{2})$. Let us verify that it works. First, $f(0) = -\frac{a}{2}$ and $f(1) = f(-1) = \frac{a}{2}$. But $f$ is an even function, so $f(-\frac{a}{2}) = f(\frac{a}{2})$ and hence $f(f(1)) = f(f(0)) = f(f(-1))$ as required.

Case 2: $f(1) = f(0)$.
This gives $b = -a$, so $f(1) = f(0) = c$ and $f(-1) = 2a + c$. Therefore
$$0 = f(f(-1)) - f(f(1)) = (f(-1) - f(1))(a(f(-1) + f(1)) + b) \\ = 2a(a(2a + 2c) - a) = 2a^2(2a + 2c - 1).$$
So $2a + 2c - 1 = 0$, and hence $c = -a + \frac{1}{2}$. This yields $f(x) = a(x^2 - x - 1) + \frac{1}{2}$. Let us verify that it works. Note that $f(\frac{1}{2} - x) = f(\frac{1}{2} + x)$. Also $f(0) = \frac{1}{2} - \frac{a}{2} = f(1)$ and $f(-1) = \frac{1}{2} + \frac{a}{2}$. So $f(f(1)) = f(f(0)) = f(f(-1))$ as required.

Case 3: $f(-1) = f(0)$.
This gives $b = a$, so $f(-1) = f(0) = c$ and $f(1) = 2a + c$. Therefore
$$0 = f(f(1)) - f(f(-1)) = (f(1) - f(-1))(a(f(1) + f(-1)) + b) \\ = 2a^2(2a + 2c + 1).$$
So $2a + 2c + 1 = 0$ and hence $c = -a - \frac{1}{2}$. This yields $f(x) = a(x^2 + x - 1) - \frac{1}{2}$. Let us verify that it works. Note that $f(-\frac{1}{2} - x) = f(-\frac{1}{2} + x)$. Also $f(0) = -\frac{a}{2} - \frac{1}{2} = f(-1)$ and $f(1) = \frac{a}{2} - \frac{1}{2}$. So $f(f(1)) = f(f(0)) = f(f(-1))$ as required.

In summary, the solutions are
$$a(x^2 - \frac{1}{2}), \quad a(x^2 - x - 1) + \frac{1}{2}, \quad a(x^2 + x - 1) - \frac{1}{2}$$

Solution 2. First note that $f(f(x))$ is a quartic. The quartic $g(x) = a x^4 + \beta x^3 + \gamma x^2 + \delta x + \epsilon$ has the property that $g(1) = g(0) = g(-1)$ if and only if $\alpha + \gamma = 0$ and $\beta + \delta = 0$. For the given function
$$f(f(x)) = a(a x^2 + b x + c)^2 + b(a x^2 + b x + c) + c$$
So the property $f(f(1)) = f(f(0)) = f(f(-1))$ gives $a^3 + a b^2 + 2 a^2 c + a b = 0$ and $2 a^2 b + 2 a b c + b^2 = 0$.
Since $a \neq 0$, the first equation gives $b^2 + b + a^2 + 2 a c = 0$. The latter equation gives $b(2 a^2 + 2 a c + b) = 0$. So $b = 0$ or $b = -2 a^2 - 2 a c = -2 a(a + c)$. We look at these cases separately.

Case 1: $b = 0$.
Then $a^2 + 2 a c = 0$ and hence $c = -a/2$ (since $a \neq 0$). Then $f(x) = a(x^2 - \frac{1}{2})$. Let us verify that it works. First, $f(0) = -\frac{a}{2}$ and $f(1) = -\frac{1}{2}$. But $f$ is even, so $f(-\frac{1}{2}) = f(\frac{1}{2})$ and hence $f(f(1)) = f(f(0)) = f(f(-1))$ as required.

Case 2: $b = -2 a^2 - 2 a c = -2 a(a + c)$.
Then the first equation gives $4 a^2(a + c)^2 - 2 a^2 - 2 a c + a^2 + 2 a c = 0$. Since $a \neq 0$, this gives $(a + c)^2 = 1/4$. Thus $c = -a \pm \frac{1}{2}$ and hence $b = \mp a$. So we get two solutions $f(x) = a(x^2 - x - 1) + \frac{1}{2}$ and $f(x) = a(x^2 + x - 1) - \frac{1}{2}$, for any $a \neq 0$.
We check that both of these work. For the first function, $f(\frac{1}{2} - x) = f(\frac{1}{2} + x)$. Also, $f(0) = \frac{1}{2} - \frac{1}{2} = \frac{1}{2}$ and $f(-1) = \frac{1}{2} + \frac{1}{2}$. So $f(f(1)) = f(f(0)) = f(f(-1))$ as required.
For the second function, $f(-\frac{1}{2} - x) = f(-\frac{1}{2} + x)$. Also, $f(0) = -\frac{1}{2} - \frac{1}{2} = f(-1)$ and $f(1) = \frac{1}{2} - \frac{1}{2}$. So $f(f(1)) = f(f(0)) = f(f(-1))$ as required.

In summary, the solutions are
$$a(x^2 - \frac{1}{2}), \quad a(x^2 - x - 1) + \frac{1}{2}, \quad a(x^2 + x - 1) - \frac{1}{2},$$
for any $a \neq 0$.