Each of the $ x^2 $-terms in the expansion of the product is obtained by multiplying the $ x $-terms from two of the 15 factors of the product. The coefficient of the $ x^2 $-term is therefore the sum of the products of each pair of numbers in the set $ \{-1,2,-3,\ldots,14,-15\} $. Note that, in general, $$(a_1+a_2+\cdots+a_n)^2=a_1^2+a_2^2+\cdots+a_n^2+2\cdot\left(\sum_{1\le i<j\le n}a_ia_j\right).$$Thus, the coefficient of $ x^2 $ is \begin{align*} \sum_{1\le i<j\le15}(-1)^{i}i(-1)^{j}j&= \frac{1}{2}\left(\left(\sum^{15}_{k=1}(-1)^{k}k\right)^2- \sum^{15}_{k=1}k^2\right)\cr &=\frac{1}{2}\left((-8)^2-\frac{15(15+1)(2\cdot15+1)}{6}\right)=-588.\cr \end{align*}$$\textbf{OR}$$Let $ C $ be the coefficient of $ x^2 $. Then \begin{align*} f(x)&=(1-x)(1+2x)(1-3x)\dotsm(1-15x)\cr &=1+(-1+2-3+\cdots-15)x+Cx^2+\cdots\cr &=1-8x+Cx^2+\cdots.\cr \end{align*}Thus $ f(-x)=1+8x+Cx^2-\cdots\, $. But $ f(-x)=(1+x)(1-2x)(1+3x)\ldots(1+15x) $, so \begin{align*} f(x)f(-x)&= (1-x^2)(1-4x^2)(1-9x^2)\dotsm(1-225x^2)\cr&= 1-(1^2+2^2+3^2+\cdots+15^2)x^2+\cdots.\end{align*}Also $ f(x)f(-x)= (1-8x+Cx^2+\cdots)(1+8x+Cx^2-\cdots)=1+(2C-64)x^2+\cdots\, $. Thus $ 2C-64=-(1^2+2^2+3^3+\cdots+15^2) $, and, as above, $ C=\boxed{-588} $.