We may similarly wonder whether any partial sums in the trinomial expansion
$$(1+z+z^2)^n = \sum_{k=0}^{2n} {n \choose k}_{\!2}z^k$$
could be $0$ when $z = e^{2i\pi/3}$ cancels the whole. The answer is positive too :
$${5 \choose 2}_{\!2}z^2+{5 \choose 3}_{\!2}+{5 \choose 7}_{\!2}z+{5 \choose 8}_{\!2}z^2 = 15z^2+30+30z+15z^2 = 0$$
. $${11 \choose 0}_{\!2} + {11 \choose 4}_{\!2}z + {11 \choose 6}_{\!2} + {11 \choose 9}_{\!2} + {11 \choose 11}_{\!2}z^2 + \\{11 \choose 13}_{\!2}z + {11 \choose 16}_{\!2}z + {11 \choose 18}_{\!2} + {11 \choose 22}_{\!2}z =\\ 1+880z+4917+19855+25653z^2+19855z+4917z+880+z = 0$$
. $${13 \choose 1}_{\!2}z+{13 \choose 3}_{\!2}+{13 \choose 5}_{\!2}z^2+{13 \choose 8}_{\!2}z^2+{13 \choose 10}_{\!2}z+{13 \choose 12}_{\!2}+\\ {13 \choose 14}_{\!2}z^2+{13 \choose 16}_{\!2}z+{13 \choose 18}_{\!2}+{13 \choose 21}_{\!2}+{13 \choose 23}_{\!2}z^2+{13 \choose 25}_{\!2}z =\\ 13z + 442 + 5005z^2 + 52624z^2 + 129844z + 201643 + \\ 201643z^2 + 129844z + 52624 + 5005 + 442z^2 + 13z = 0$$