I would like to apply the hints presented in the answer by Bazin to give the following answer.

Every derivation of $\chi^{\infty}(M)$ is inner and it is obvious that every inner derivation is a non elliptic operator. The reason is that every non vanishing vector field is locally look like $\partial/\partial_x$, whose corresponding adjoint operator is obviousely non elliptic.

So the non ellipticity of a derivation operator lies in its adjoint-ness not merely on its order.It is first order but this order situation is not enough to conclude it is non elliptic. In fact there is an example of a first order elliptic pde in dim 3 as follows:
$$D(u,v, w)= (u_x-v_y, u_y+v_x, w)$$

This operator $D$ satisfies the definition of an elliptic PDE but is of first order.

One can construct a first order elliptic PDE in dimension 4 without a term of order 0. Construct a linear PDE on $\chi^{\infty}(\mathbb{R}^4)$ whose principal symbol is the $4\times 4$ matrix representation of quaterniouns $$\xi_1+\xi_2 i + \xi_3 j+\xi_4 k$$ where $(\xi_1,\xi_2,\xi_3,\xi_4)$ is a cotangent vector.

So it seems that the first line of the answer by Bazin does not work.

**Remark:** The above linked paper contains a reference to a result by F.Takens that every derivation of $\chi^{\infty}(M)$ is an inner derivation.