No, there doesn't exist such set.

For contradiction assume there exists such multiset $A=\{a_1,a_2,\dots,a_n\}$. Claim: There is no element equal to $0$ in this set. Proof: Assume there is an element equal to $0$ in $A$, taking $0$ in the set of $1008$ roots of a polynomial $P$ there must be another element equal to $0$ in $A$ because $x\mid P(x)$ , now taking these two $0$ elements as roots, again, with the same argument there must be another two $0$ elements in $A$, continuing like this all elements of $A$ must be $0$, contradiction. $\blacksquare$. Since the $2016$ numbers are all non-zero there are at least $1008$ of them of the same sign(positive or negative). If there are $1008$ positive numbers in $A$, call this $1008$ element set $B\subset A$, we consider the polynomial with these $1008$ positive numbers as roots. Now, from Vieta's relations we have that $504$ of the coefficients are positive and $504$ of the coefficients are negative. Consider the product $Q$ of these $1008$ positive roots, let $a_k$ be the smallest element in $B$ and let $a_l$ be the maximal element from the $504$ negative numbers (in modulus). Considering the polynomial with roots $(B\setminus a_k)\cup a_l$ we have that $\frac{Q\cdot a_l}{a_k}=a_j$ where $a_j$ is negative. But we know that $|Q\cdot a_l| \geq |a_j\cdot a_k|$ where equality holds only if $B=\{1,1,1\dots,1\}$ and $a_j=a_l$, and with the same argument we can show that the other $504$ positive elements are equal to $1$ wich is not true since the polynomial $(x-1)^{1008}$ doesn't have $504$ coefficients equal to $1$. The case where there are $1008$ negative elements is easier because a polynomial with $1008$ negative roots has all coefficients positive and this turns into the above case.$\blacksquare$