A polynomial with integer coefficients is called "nice" if the coefficients in the terms whose degree is not a multiple of $3$, are multiples of $3$, and their sum is a multiple of $6$. Clearly, the sum of nice polynomials is a nice polynomial, and the sum of cube-represented polynomials is a cube-represented polynomial. We shall prove the following lemma:

Lemma. A polynomial is cube-represented if and only if it is nice.

Consider the cube of a polynomial $P(x)$ with integer coefficients, where:
$$P(x) = a_{n} x^{n} + \cdots + a_{1} x + a_{0}$$
After expanding, all terms of $P^{3}(x)$ are either of the form $a_{i}^{3} x^{3i}$, or $3 a_{i}^{2} a_{j} x^{2i + j}$ with $i$ different from $j$, or of the form $6 a_{i} a_{j} a_{k} x^{i + j + k}$ with $i, j, k$ distinct.

We have $3 \mid 2i + j$ if and only if $3 \mid 2j + i$, and $a_{i}^{2} a_{j} + a_{j}^{2} a_{i} = a_{i} a_{j}(a_{i} + a_{j})$ which is always an even number. From here, clearly a cube-represented polynomial has to be a nice polynomial, as defined above.

Second, consider a nice polynomial $P(x)$. We will prove by induction on $n$ that $P(x)$ is cube-represented, with $n$ being the integer such that the degree of $P(x)$ does not exceed $3n$, but is larger than $3n-3$.

For the base case where $n=1$, $P(x) = a_{3} x^{3} + 3a_{2} x^{2} + 3a_{1} x + a_{0}$, where $2 \mid a_{1} - a_{2}$.
We can write
$$P(x) = \frac{a_{1} + a_{2}}{2} (x+1)^{3} + \frac{a_{1} - a_{2}}{2} (x-1)^{3} + (a_{3} - a_{1}) x^{3} + (a_{0} - a_{2}) 1^{3}$$
which is obviously a cube-represented polynomial.

Assume the result is true for $n-1$. If $P(x) = a_{3n} x^{3n} + 3a_{3n-1} x^{3n-1} + 3a_{3n-2} x^{3n-2} + Q(x)$ is nice, where the degree of $Q(x)$ is at most $3n-3$.

Then either $a_{3n-1} + a_{3n-2}$ is even and $Q(x)$ is good, or $a_{3n-1} + a_{3n-2}$ is odd, and $Q(x) - 3x^{3n-4}$ is good.

In the first case
$$\begin{aligned} & P(x) = \frac{a_{3n-1} + a_{3n-2}}{2} (x^{n} + x^{n-1})^{3} + \frac{a_{3n-2} - a_{3n-1}}{2} (x^{n} - x^{n-1})^{3} + \\ & (a_{3n} - a_{3n-2}) x^{3n} + Q(x) - a_{2} (x^{n-1})^{3} \end{aligned}$$

In the second case
$$\begin{aligned} P(x) &= \frac{a_{3n-1} + a_{3n-2} - 1}{2} (x^{n} + x^{n-1})^{3} + \frac{a_{3n-2} - a_{3n-1} - 1}{2} (x^{n} - x^{n-1})^{3} \\ & + (x^{n} + x^{n-2})^{3} + (a_{3n} - a_{3n-2}) x^{3n} + (Q(x) - 3x^{3n-4}) - a_{2} x^{3n-3} \end{aligned}$$
so in both cases, we have $P(x)$ is a nice polynomial. Our lemma has been proved.

Back to our problem,

1. With the lemma, we can easily verify that $P(x) = 3x^{2}$ is not a nice polynomial, so it cannot be a cube-represented polynomial.

2. We need to count the number of triples $(a, b, c)$ which belong to the set $\{1,2, \ldots, 2017\}$ such that $P(x) = a x^{2} + b x + c$ is cube-represented, that is, it's nice.

$P(x)$ is nice if and only if $a, b$ are multiples of $3$ and $a + b$ is a multiple of $2$.

We have $a, b \in \{3, 6, \ldots, 2016\}$ and $a + b$ is even, there are $336^{2}$ such pairs. Value $c$ can be arbitrary, so the number of triples $(a, b, c)$ is $2017 \cdot 336^{2}$, which is the number of quadratic cube-represented polynomials satisfying the problem.