B
解:∵f(x)=-2x3-x,
∴f(-x)=2x3+x=-(-2x3-x)=-f(x),
∴函数f(x)是奇函数,且f(x)=-2x3-x在R上为减函数,
∵x1+x2>0,x2+x3>0,x3+x1>0,
∴x1>-x2,x2>-x3,x3>-x1,
则f(x1)<f(-x2),f(x2)<f(-x3),f(x3)<f(-x1),
即f(x1)<-f(x2),f(x2)<-f(x3),f(x3)<-f(x1),
∴不等式两边相加得f(x1)+f(x2)+f(x3)<-[f(x1)+f(x2)+f(x3)],
即f(x1)+f(x2)+f(x3)<0.
故选:B.