Compute $AC = 7\sqrt{3}$ and $AD = 13\sqrt{3};$ triangles $ABC$ and $AED$ turn out to have right angles at $C$ and $D$

Split $f(X) = (BX + EX)$ $+ (AX + CX + DX)$ and show that the Fermat point of triangle $ACD$ lies on segment $BE$

In triangle $ABC,$ the law of cosines with $\angle B = 60^\circ$ gives $AC^2 = 14^2 + 7^2 - 14 \cdot 7 = 147,$ so $AC = 7\sqrt{3};$ since $7^2 + 147 = 14^2,$ the angle at $C$ is right and $\angle BAC = 30^\circ.$ Likewise $AD = 13\sqrt{3},$ with a right angle at $D$ and $\angle DAE = 30^\circ.$ In triangle $ACD$ with $CD = 24,$ $$\begin{aligned} \cos \angle CAD &= \frac{147 + 507 - 576}{2 \cdot 7\sqrt{3} \cdot 13\sqrt{3}} \\ &= \frac{1}{7}, \\ \sin \angle CAD &= \frac{4\sqrt{3}}{7}. \end{aligned}$$

Split $f(X) = (BX + EX)$ $+ (AX + CX + DX) \ge BE + T,$ where $T$ is the minimum of $AX + CX + DX.$ Since $\angle BAE = 30^\circ + \angle CAD + 30^\circ,$ we get $\cos \angle BAE = \frac{1}{2} \cdot \frac{1}{7}$ $- \frac{\sqrt{3}}{2} \cdot \frac{4\sqrt{3}}{7}$ $= -\frac{11}{14},$ so $BE^2 = 14^2 + 26^2$ $+ 2 \cdot 14 \cdot 26 \cdot \frac{11}{14} = 1444$ and $BE = 38.$ All angles of triangle $ACD$ are less than $120^\circ,$ so $T$ is attained at its Fermat point; erecting an equilateral triangle $ACP$ on side $AC$ away from $D,$ the standard rotation argument gives $T = PD,$ and since $\angle PAD = 60^\circ + \angle CAD$ also has cosine $-\frac{11}{14},$ $$\begin{gathered} T^2 \\ = 147 + 507 \\ {}+ 2 \cdot 7\sqrt{3} \cdot 13\sqrt{3} \cdot \frac{11}{14} \\ = 1083, \\ T = 19\sqrt{3}. \end{gathered}$$

Both bounds are tight simultaneously: let $F$ be the Fermat point of $ACD,$ so $\angle AFC = \angle AFD = 120^\circ.$ Since $\angle AFC + \angle ABC = 180^\circ,$ point $F$ lies on the circumcircle of $ABC,$ whence $\angle AFB = \angle ACB = 90^\circ;$ similarly $F$ lies on the circumcircle of $AED$ and $\angle AFE = \angle ADE = 90^\circ.$ Thus $\angle BFE = 180^\circ,$ so $F$ lies on segment $BE$ and $f(F) = BE + T = 38 + 19\sqrt{3}.$ The answer is $m + n + p = 38 + 19 + 3 = 60.$