如图，作点$A$关于$BE$的对称点${A\’}$，连接$BA\’$，$AE$，$CA\’$.

$\because \triangle EBD$是等边三角形，
$\therefore \angle EBD=60^{\circ}$，$BE=BD$，
$\because AB=AC$，$\angle BAC=120^{\circ}$，
$\therefore \angle ABC=\angle ACB=30^{\circ}$，
$\therefore \angle ABE=\angle ABD=30^{\circ}$，
$\because BE=BD$，$BA=BA$，
$\therefore \triangle ABE$≌$\triangle ABD\left(SAS\right)$，
$\therefore AE=AD$，
$\therefore \angle CBA\’=90^{\circ}$，$CA\’=2BA\’=4\sqrt{3}$.
$\because EA=EA\’$，
$\therefore AE+EC=EA\’+EC\geqslant CA\’$，
$\therefore AE+EC\geqslant 4\sqrt{3}$，
$AD+CE$的最小值为$4\sqrt{3}$.
故答案为：$4\sqrt{3}$.

如图，作点$A$关于$BE$的对称点${A\’}$，连接$BA\’$，$AE$，$CA\’$.

$\because \triangle EBD$是等边三角形，
$\therefore \angle EBD=60^{\circ}$，$BE=BD$，
$\because AB=AC$，$\angle BAC=120^{\circ}$，
$\therefore \angle ABC=\angle ACB=30^{\circ}$，
$\therefore \angle ABE=\angle ABD=30^{\circ}$，
$\because BE=BD$，$BA=BA$，
$\therefore \triangle ABE$≌$\triangle ABD\left(SAS\right)$，
$\therefore AE=AD$，
$\therefore \angle CBA\’=90^{\circ}$，$CA\’=2BA\’=4\sqrt{3}$.
$\because EA=EA\’$，
$\therefore AE+EC=EA\’+EC\geqslant CA\’$，
$\therefore AE+EC\geqslant 4\sqrt{3}$，
$AD+CE$的最小值为$4\sqrt{3}$.
故答案为：$4\sqrt{3}$.