Assuming that the OP meant "the intersection $L^1L^1\cap L^1L^\infty$", not the Minkowsi sum:

For simplicity let me define $\beta(s):=\|F(s,\cdot)/(1+|.|)\|_{\infty}\in L^1(0,T)$ and $B_T:=\int_0^T\beta(s)ds=\|F/(1+|x|)\|_{L^1L^\infty}$. Writing \begin{multline*} |\Phi(t,x)-x|\leq \int_0^t |F(s,\Phi(s,x))|ds\\ \leq \int_0^t\frac{|F(s,\Phi(s,x))|}{1+|\Phi(s,x)|}(1+|\Phi(s,x)|)ds \leq \int_0^t \beta(s) (1+|\Phi(s,x)|)ds \end{multline*} we get, with $|\Phi(s,x)|\leq |x|+|\Phi(s,x)-x|\leq R+|\Phi(s,x)-x|$, $$ |\Phi(t,x)-x|\leq (1+R)\int_0^t\beta(s)ds +\int_0^t\beta(s)|\Phi(s,x)-x|ds. $$ Applying Grönwall's inequality in its integral form, you readily conclude that $$ |\Phi(t,x)-x|\leq (1+R)\int_0^t\beta(s)ds \exp\left(\int_0^t\beta(s)ds\right))\leq (1+R)B_T\exp(B_T) $$ and this gives you the "truncated cone" C(T).

Edited according to Martin Hairer's comment: the flow $\Phi(t,0)$ can blow up in finite (and arbitrarily small) time if the $L^1(0,T;L^1)$ component $F_1\neq 0$ in the Minkowski sum $\frac{F}{(1+|x|)}= F_1+F_2\in L^1(0,T;L^1) + L^1(0,T;L^\infty)$. So with the OP's assumption there is no hope for a reasonable answer, hence from now on we simply assume that $$ \frac{F}{1+|x|}\in L^1(0,T;L^\infty). $$

For simplicity let me define $\beta(s):=\|F(s,\cdot)/(1+|.|)\|_{\infty}\in L^1(0,T)$ and $B_T:=\int_0^T\beta(s)ds=\|F/(1+|x|)\|_{L^1L^\infty}$. Writing \begin{multline*} |\Phi(t,x)-x|\leq \int_0^t |F(s,\Phi(s,x))|ds\\ \leq \int_0^t\frac{|F(s,\Phi(s,x))|}{1+|\Phi(s,x)|}(1+|\Phi(s,x)|)ds \leq \int_0^t \beta(s) (1+|\Phi(s,x)|)ds \end{multline*} we get, with $|\Phi(s,x)|\leq |x|+|\Phi(s,x)-x|\leq R+|\Phi(s,x)-x|$, $$ |\Phi(t,x)-x|\leq (1+R)\int_0^t\beta(s)ds +\int_0^t\beta(s)|\Phi(s,x)-x|ds. $$ Applying Grönwall's inequality in its integral form (and observing that $t\mapsto\int_0^t\beta(s)ds$ is continuous nondecreasing), we can conclude that $$ |\Phi(t,x)-x|\leq (1+R)\int_0^t\beta(s)ds \exp\left(\int_0^t\beta(s)ds\right))\leq (1+R)B_T\exp(B_T) $$ and this gives the "truncated cone" $C(T)$.