The value of unioncheck[n] can be derived as
$\frac{n^n}{n!}-1$.
$p_0$, The value of the first permutation (i.e. Range[n]) is given by
$$p_0 = 1+ \underset{i=0}{\overset{n-1}{\sum }}i n^{n-i-1} $$
while the
$p_{n!}$, the value of the last permutation (i.e. Reverse@Range[n]) is given by
$$p_{n!} = 1+ \underset{i=0}{\overset{n-1}{\sum }}i n^i$$
The sum of the two is given by
$$ p_0 + p_{n!} = \underset{i=0}{\overset{n-1}{\sum }}i \left(n^i+n^{-i+n-1}\right)+2 \\
= n^n+1$$
The sum of the first and last values of reflocs[n] is
$r = (n!+1) \frac{n^n}{n!} $, so therefore the value of unioncheck[n] is given by
$$r- ( p_0 + p_{n!} )= (n!+1) \frac{n^n}{n!} - ( n^n+1 )= \boxed{\frac{n^n}{n!} - 1}$$
Table[n^n/n! - 1,{n,7}]
(*{0, 1, 7/2, 29/3, 601/24, 319/5, 116929/720}*)
You may also be interested in the fact that the position of a permutation perm of Range[n] in Tuples[Range[n],n] is given exactly by 1+FromDigits[perm -1,n] :
Flatten@Table[permlocs[n], {n, 5}] ==
Flatten@Table[
1 + FromDigits[# - 1, n] & /@ Permutations[Range@n], {n, 5}]
(*True*)
See also OEIS sequences A134640, A058128 and A062813
