There are $\binom53$ ways to choose which $3$ candidates come equal first. Since $5\cdot8=40$, their score must be at least $9$, and it can obviously be at most $13$. Now make a little table:
$\begin{array}{c|l} \text{Top vote}&\text{Remaining votes}\\ \hline 9&8,5;\quad7,6\\ 10&9,1;\quad8,2;\quad7,3;\quad6,4;\quad5,5\\ 11&7,0;\quad6,1;\quad5,2;\quad4,3\\ 12&4,0;\quad3,1;\quad2,2\\ 13&1,0 \end{array}$
If the top vote is $9$, the remaining votes must have been split either $8,5$ or $7,6$. That’s $2$ possibilities, and each can happen in $2$ ways, depending on which of the remaining two voters gets the larger number of votes. Thus, there are $\binom53\cdot2\cdot2=40$ ways to get the vote totals $9,9,9,8,5$ and $9,9,9,7,6$, corresponding to the $40$ distinguishable permutations of those five numbers.
If the top vote is $11$, there are $4$ different ways to split the remaining $7$ votes into two parts, and each of those can be assigned to the other two candidates in either order, so there are $\binom53\cdot4\cdot2=80$ ways to get three vote totals of $11$ and two more adding up to $7$.
That’s two of the five cases; the other three can be done similarly, so I’ll leave them for you to try.