Probability to obtain more than X with 3 dice.

Your concrete problem is already solved here: If we throw three dice.

The general problem is equivalent to counting the number of ways of distributing $X-Y$ balls into $Y$ bins with limited capacity $Z-1$. This problem is solved at Balls In Bins With Limited Capacity using inclusion-exclusion. The result is

$$ \sum_{t=0}^Y(-1)^t\binom Yt\binom{X-tZ-1}{Y-1}\;, $$

where, contrary to convention, the binomial coefficient is taken to be zero for negative upper index.

This is the count of outcomes with sum exactly $X$; to get the probability of a sum of more than $X$, we need to sum from $X+1$ to $YZ$ and divide by the number $Z^Y$ of equiprobable outcomes:

$$ Z^{-Y}\sum_{x=X+1}^{YZ}\sum_{t=0}^Y(-1)^t\binom Yt\binom{x-tZ-1}{Y-1}=Z^{-Y}\sum_{t=0}^Y(-1)^t\binom Yt\left(\binom{YZ-tZ}Y-\binom{X-tZ}Y\right)\;. $$

For $Y=3$, $Z=6$, this is

\begin{align} &\frac1{216}\sum_{t=0}^3(-1)^t\binom 3t\left(\binom{18-6t}3-\binom{X-6t}3\right)\\ ={}&\frac1{216}\left(\binom{18}3-\binom X3-3\left(\binom{12}3-\binom{X-6}3\right)+3\left(\binom63-\binom{X-12}3\right)\right)\\ ={}&1-\frac1{216}\left(\binom X3-3\binom{X-6}3+3\binom{X-12}3\right)\;, \end{align}

where again binomial coefficients with negative upper index are taken to be zero. Distinguishing the three cases, we can write this as

$$ \frac1{1296}\begin{cases} -X^3+3X^2-2X+1296&3\le X\lt9\;,\\ 2X^3-60X^2+436X+288&6\le X\lt15\;,\\ -X^3+57X^2-1082X+6840&12\le X\le18 \end{cases} $$

(where I intentionally wrote the maximal overlapping ranges to exhibit the symmetry more clearly). As far as I checked, the results coincide with those of the concrete calculation linked to above.