C

如图，延长$AF$，$BC$交于点$H$，

$\because AD\bot AB$，$BC\bot AB$，
$\therefore AD$∥$BC,\angle ABC=\angle BAD=90^{\circ}$，
$\therefore \angle DAF=\angle H$，
$\because$点$F$是$CD$中点，
$\therefore CF=DF$，
又$\because \angle AFD=\angle CFH$，
$\therefore \triangle ADF$≌$\triangle HCF\left(AAS\right)$，
$\therefore AF=FH$，$AD=CH$，
$\because AE=AD$，$BE=BC$，
$\therefore AE+BE=BC+AD=BC+CH$，
$\therefore AB=BH$，
又$\because AF=FH$，$\angle ABC=90^{\circ}$，
$\therefore AF=FH=BH$，$AF\bot BF$，故①正确；
$\because AF=FH$，
$\therefore S_{\triangle ABF}=S_{\triangle BFH}$，
$\because \triangle ADF$≌$\triangle HCF$，
$\therefore S_{\triangle ADF}=S_{\triangle FCH}$，
$\therefore S_{\triangle ADF}+S_{\triangle BCF}=S_{\triangle AFB}$；故②正确；
$\because AE=AD=3$，$AB=11$，
$\therefore BE=8$，
$\because AD\bot AB$，$BC\bot AB$，$AE=AD=3$，$BE=BC=8$，
$\therefore \angle DEA=\angle ADE=45^{\circ}=\angle BEC=\angle BCE$，$DE=3\sqrt{2}$，$CE=8\sqrt{2}$，
$\therefore \angle DEC=90^{\circ}$，
$\therefore DC^{2}=DE^{2}+EC^{2}=18+128=146$，故③错误；
$\because S_{\triangle ADE}=\frac{1}{2}AD^{2}$，$S_{\triangle BEC}=\frac{1}{2}BC^{2}$，$S_{\triangle DEC}=\frac{1}{2}\times DE\cdot EC=\frac{1}{2}\times \sqrt{2}AD\times \sqrt{2}BC=AD\cdot BC$，
$\therefore S_{\triangle ADE}+S_{\triangle BCE}=\frac{1}{2}(AD^{2}+BC^{2})$，
$\because \left(BC-AD\right)^{2}\geqslant 0$，
$\therefore BC^{2}+AD^{2}\geqslant 2AD\cdot BC$，
$\therefore \frac{1}{2}(AD^{2}+BC^{2})\geqslant AD\cdot BC$，
$\therefore S_{\triangle ADE}+S_{\triangle BEC}\geqslant S_{\triangle DEC}$，故④正确；
故选：$C$.