From equal tangents we have congruent triangles and so by averaging the sum is $360^\circ$. P.S. : I really like the "rotation" in the angle chase we did!

Excellent problem! Answer: $\boxed{360^{\circ}}$. Let $G$ be the contact point of the insphere with face $ABC$. Notice that $\triangle SDA$ and $\triangle SFA$ are congruent, since they have equal corresponding sides. Also, $\triangle AFC$ and $\triangle AGC$ are congruent so $\measuredangle AFC=\measuredangle AGC$. Similar relations show that $$\measuredangle AFC+\measuredangle CEB+\measuredangle BDA=\measuredangle AGC+\measuredangle CGB+\measuredangle BGA=360^{\circ}.$$Hence, we see that $$\measuredangle SDA+\measuredangle SFC=360^{\circ}-\measuredangle AFC.$$Adding cyclically obtained relations, we conclude that $$2(\measuredangle SDA+\measuredangle SEB+\measuredangle SFC)=1080^{\circ}-(\measuredangle AFC+\measuredangle CEB+\measuredangle BDA)=720^{\circ}.$$

Let sphere tangent to $ABC$ by $G.$ From $SDA\cong SFA$ and all similar assertions we obtain $$\angle SDA+\angle SEB+\angle SFC=\frac{1}{2}(6\pi-(\angle AGB+\angle BGC+\angle CGA))=2\pi.$$