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).