I'm really not sure on how to tackle this. I have tried considering all matrices comprised of three of the five vectors -- as rows -- and row reducing them, but that hasn't really lead anywhere (so far).
I know that the cone consists of all conic combinations of the generating vectors. I am unsure of how to glean its irredundant inequalities.
We have $5$ points $A,B,C,D,E,$ such that every $4$ points determine an affine hyperplane. Let's describe the one corresponding to $A,B,C$ and $D,$ i.e. we discard $E.$ We form $3$ vectors $v=\vec{AB},u=\vec{AC},w= \vec{AD}.$ The equation of the hyperplane is given by $$\begin{vmatrix} {\bf x_1} & \bf{x_2} & \bf x_3 & \bf x_4\\ u_1 & u_2 & u_3 & u_4 \\ v_1 & v_2 & v_3 & v_4 \\ w_1 & w_2 & w_3 & w_4 \end{vmatrix}=0,$$ while the affine hyperplane is described by $$\begin{vmatrix} {\bf x_1} & \bf{x_2} & \bf x_3 & \bf x_4\\ u_1 & u_2 & u_3 & u_4 \\ v_1 & v_2 & v_3 & v_4 \\ w_1 & w_2 & w_3 & w_4 \end{vmatrix} =\begin{vmatrix} a_1 & a_2 & a_3 & a_4\\ u_1 & u_2 & u_3 & u_4 \\ v_1 & v_2 & v_3 & v_4 \\ w_1 & w_2 & w_3 & w_4 \end{vmatrix}=:\Delta_A,$$ where $A=(a_1,a_2,a_3,a_4).$ Then for $E=(e_1,e_2,e_3,e_4)$ we calculate the determinant $$ \Delta_E:=\begin{vmatrix} e_1 & e_2 & e_3 & e_4\\ u_1 & u_2 & u_3 & u_4 \\ v_1 & v_2 & v_3 & v_4 \\ w_1 & w_2 & w_3 & w_4 \end{vmatrix}.$$ If $\Delta_E>\Delta_A,$ one of the inequalities describing the region is of the form $$\begin{vmatrix} {\bf x_1} & \bf{x_2} & \bf x_3 & \bf x_4\\ u_1 & u_2 & u_3 & u_4 \\ v_1 & v_2 & v_3 & v_4 \\ w_1 & w_2 & w_3 & w_4 \end{vmatrix}\ge \begin{vmatrix} a_1 & a_2 & a_3 & a_4\\ u_1 & u_2 & u_3 & u_4 \\ v_1 & v_2 & v_3 & v_4 \\ w_1 & w_2 & w_3 & w_4 \end{vmatrix}.$$ We repeat that procedure for $4$ remaining cases and obtain $5$ inequalities describing the region.