The final answer is
9
To figure out the digits (building off some of the logic in Earlien's solution):
$N=I^2$ means $I$=2 or 3 and $N$=4 or 9. 0 or 1 are not possible if $N$ and $I$ are to be unique.
Since $N$ = 4 or 9, $N^M=A$ means $M$=0 and $A$=1 ($M$=1 will result in non-uniqueness, and larger $M$ result in $A$>9).
$I^S=D$ means ${I,S,D}$ = {3,2,9} or {2,3,8} since 0 and 1 are taken. However, $I$=3 and $D$=9 is impossible because $N$=9 when $I$=3. So now we can confirm $I$=2, $N$=4, $D$=8, and $S$=3.
$N! = C(T-O)^I$ simplifies to $24/C = (T-O)^2$. $C$=6 because that is the only possibility that gives a perfect square. Then $T$=9 or 7 and $O$=7 or 5.
Since the vowels must add up to a prime number, $E$=3, 5, or 9 if $O$=5, or $E$=1, 3, 7, or 9 if $O$=7. Since all those digits are taken except for $E$=9, then $O$=5, $E$=9, and $T$=7. We now have all the digits we need.
To figure out the missing number:
My first instinct was also to subtract the inputs from each other. However, this fails in the last step, where 20-10 != 11.
Instead, it seems you add all the digits of the inputs, so 1+3+2+3 = 9. Incredible how most of the numbers satisfy both patterns, until the last step befuddles you! Honestly, if I wasn't thinking about digits already, this would have been even tougher.
Upon some searching it seems this puzzle is inspired by:
Nob's number puzzle, a famous and similar number tree puzzle designed by Japanese puzzle guru Nobuyuki Yoshigahara.
Very clever, thanks for the puzzle!