Answer: 588

Small Hint:

The $x^2$ coefficient is the sum of the products of pairs from $-1, 2, -3, \ldots, 14, -15$

Big Hint:

Use $2\sum_{i \lt j} a_i a_j = \left(\sum a_i\right)^2 - \sum a_i^2,$ and the alternating sum is $-8$

Solution:

Write the product as $\prod_{k=1}^{15} (1 + a_k x)$ with $a_k = (-1)^k k.$ An $x^2$ term arises by choosing the $x$-term from two factors, so $C = \sum_{i \lt j} a_i a_j,$ and $$C = \frac{\left(\sum a_k\right)^2 - \sum a_k^2}{2}.$$

The alternating sum is $$\begin{aligned} &(-1 + 2) + (-3 + 4) \\ &\quad {}+ \cdots + (-13 + 14) - 15 \\ &= 7 - 15 \\ &= -8, \end{aligned}$$ and $$\begin{aligned} \sum a_k^2 &= 1^2 + 2^2 + \cdots + 15^2 \\ &= \frac{15 \cdot 16 \cdot 31}{6} \\ &= 1240. \end{aligned}$$

Thus $C = \frac{64 - 1240}{2} = -588,$ so $|C| = 588.$