Let the total on the three six-sided dice be $X,$ $3\leq X\leq18.$
Let the number on the twenty-sided die be $Y,$ $1\leq Y\leq20.$
Given any particular value of $X,$ the probability
that the twenty-sided die will roll higher is
$$P(Y>X \mid X) = \frac{20-X}{20} = 1 - \frac1{20}X.$$
The overall probability that the twenty-sided die will roll higher than
the total on the other three dice is
\begin{align}
P(Y>X) &= \sum_{n=3}^{18} P(Y > X \mid X)P(X=n) \\
&= \sum_{n=3}^{18} \left(1 - \frac1{20}X\right)P(X=n).
\end{align}
The last line of that set of equations is just the expected value of
$1 - \frac1{20}X.$
That is,
\begin{align}
P(Y>X) &= \mathbb E\left[1 - \frac1{20}X\right] \\
&= 1 - \frac1{20} \mathbb E[X] \\
&= 1 - \frac1{20} \left(\frac{21}{2}\right) \\
&= \frac{19}{40}.
\end{align}
If the question is actually the one posed in the original question body rather than in the original title,
namely the probability that $X > Y,$ then we simply observe that
for any given value of $X,$
$$ P(Y < X \mid X) = \frac{X-1}{20} = \frac{1}{20}X - \frac{1}{20}.$$
The rest of the calculation builds on this the same way the first calculation in this answer built on $P(Y > X \mid X).$
We find that
\begin{align}
P(Y<X) &= \mathbb E\left[\frac1{20}X - \frac1{20}\right] \\
&= \frac1{20} \mathbb E[X] - \frac1{20}\\
&= \frac1{20} \left(\frac{21}{2}\right) - \frac1{20} \\
&= \frac{19}{40}.
\end{align}
This should not be surprising, because it also follows from
$P(Y>X)=\frac{19}{40}$ and the "obvious" fact that $P(Y=X)=\frac1{20}.$
The probability that all sides are the same is $6$ times the probability that all sides are $1$. So we don’t have to worry about the distribution of all numbers on the die, we just have to keep track of the number of $1$s. That yields a Markov chain with $7$ different states, two of which ($0$ and $6$) are absorbing. The transition matrix is
$$
\mathsf P(i\to j)=6^{-6}\binom6ji^j(6-i)^{6-j}
$$
(where $0^0=1$), or in matrix form:
$$
P=6^{-6}\pmatrix{
46656&0&0&0&0&0&0\\
15625&18750&9375&2500&375&30&1\\
4096&12288&15360&10240&3840&768&64\\
729&4374&10935&14580&10935&4374&729\\
64&768&3840&10240&15360&12288&4096\\
1&30&375&2500&9375&18750&15625\\
0&0&0&0&0&0&46656\\
}\;.
$$
To my surprise, this matrix has a rather nice eigensystem:
$$
P=\pmatrix{6\\5&5&-5&25&-5&3725&1\\4&8&-4&-8&8&-10576&2\\3&9&0&-27&0&14337&3\\2&8&4&-8&-8&-10576&4\\1&5&5&25&5&3725&5\\&&&&&&6}\\
\times\pmatrix{6\\&38160\\&&120\\&&&2448\\&&&&120\\&&&&&648720\\&&&&&&6}^{-1}
\\
\times
\pmatrix{1\\&\frac56\\&&\frac59\\&&&\frac5{18}\\&&&&\frac5{54}\\&&&&&\frac5{324}\\&&&&&&1}
\\\times\pmatrix{1\\-2681&981&1125&1150&1125&981&-2681\\7&-8&-5&0&5&8&-7\\-14&33&-6&-26&-6&33&-14\\1&-4&5&0&-5&4&-1\\-1&6&-15&20&-15&6&-1\\&&&&&&1}\;,
$$
where the first diagonal matrix is just for normalization (so that I could write the matrices containing the left and right eigenvectors with integers) and the second diagonal matrix contains the eigenvalues.
Thus, since we start in state $1$, the probability to have reached state $6$ after $n$ rolls is
$$
-\frac{5\cdot2681}{38160}\left(\frac56\right)^n+\frac{5\cdot7}{120}\left(\frac59\right)^n-\frac{25\cdot14}{2448}\left(\frac5{18}\right)^n+\frac{5\cdot1}{120}\left(\frac5{54}\right)^n-\frac{3725\cdot1}{648720}\left(\frac5{324}\right)^n+\frac16
\\[3pt]
=
-\frac{2681}{7632}\left(\frac56\right)^n+\frac7{24}\left(\frac59\right)^n-\frac{175}{1224}\left(\frac5{18}\right)^n+\frac1{24}\left(\frac5{54}\right)^n-\frac{745}{129744}\left(\frac5{324}\right)^n+\frac16\;.
$$
We need to multiply this by $6$ to get the probability of having reached this state for any of the $6$ numbers on the die in order to get the probablity that the number $N$ of required rolls is less than or equal to $n$:
$$
\mathsf P(N\le n)=1-\frac{2681}{1272}\left(\frac56\right)^n+\frac74\left(\frac59\right)^n-\frac{175}{204}\left(\frac5{18}\right)^n+\frac14\left(\frac5{54}\right)^n-\frac{745}{21624}\left(\frac5{324}\right)^n\;.
$$
Here we can check that $\mathsf P(N\le0)=0$ and $\mathsf P(N\le1)=6^{-5}$, as they must be. The probability that we need exactly $n$ rolls is
\begin{eqnarray}
\mathsf P(N=n)
&=&
\mathsf P(N\le n)-\mathsf P(N\le n-1)
\\[3pt]
&=&
\frac{2681}{6360}\left(\frac56\right)^n-\frac75\left(\frac59\right)^n+\frac{455}{204}\left(\frac5{18}\right)^n-\frac{49}{20}\left(\frac5{54}\right)^n+\frac{47531}{21624}\left(\frac5{324}\right)^n
\end{eqnarray}
for $n\gt0$ and $\mathsf P(N=0)=0$. Here’s a plot, in agreement with your numerical results. The expected value of $N$ is
\begin{eqnarray}
\mathsf E[N]
&=&\sum_{n=0}^\infty\mathsf P(N\gt n)
\\[3pt]
&=&
\sum_{n=0}^\infty\left(\frac{2681}{1272}\left(\frac56\right)^n-\frac74\left(\frac59\right)^n+\frac{175}{204}\left(\frac5{18}\right)^n-\frac14\left(\frac5{54}\right)^n+\frac{745}{21624}\left(\frac5{324}\right)^n\right)
\\[6pt]
&=&
\frac{2681}{1272}\cdot\frac6{6-5}-\frac74\cdot\frac9{9-5}+\frac{175}{204}\cdot\frac{18}{18-5}-\frac14\cdot\frac{54}{54-5}+\frac{745}{21624}\cdot\frac{324}{324-5}
\\[6pt]
&=&
\frac{31394023}{3251248}
\\[6pt]
&\approx&9.656\;,
\end{eqnarray}
also in agreement with your numerical results.
Best Answer
I'll let $P(n)$ denote the probability of $X = n$ for brevity. If you want the distribution of $X$, then it's clear that $P(2) = 1/6$, and there's a recurrence $$P(n+1) = \frac{n}{6}\left( 1 - \sum_{i=2}^n P(i) \right) $$ because the probability that the first $n$ rolls have no repeats is $1 - \sum_{i=2}^n P(i)$ and the probability that the $(n+1)$th roll repeats one of the first $n$ is $n/6$. You can tabulate the values of $P(n)$ by hand without much trouble.