You're nearly right. You've found the probability that the first three are 6 and the remaining seven are not 6. For example, you haven't taken account of $6111616111$.
Now, there are $\binom{10}{3}$ possible ways we could have got three 6s, so you need to multiply your answer by $\binom{10}{3}$.
There are ${8 \choose 2}$ ways to place the threes, then ${6 \choose 3}$ ways to place the ones, and then ${3 \choose 3}$ to place the remaining sixes.
All together we can arrange these groups in
$${8\choose 2}\cdot{6\choose3}=\frac{8!}{2!\cdot6!}\cdot\frac{6!}{3!\cdot3!}=\frac{8!}{2!\cdot3!\cdot3!}$$
different ways.
Notice that this is a multinomial distribution which takes the form
$$P(X_1=x_1,...,X_k=x_k)=\frac{n!}{x_1!\cdots x_k!}p_1^{x_1}\cdots p_k^{x_k}$$
Let $X_3,X_1,X_6$ denote the numbers of threes, ones, and sixes observed, respectively. Then
$$P(X_3=2, X_1=3,X_6=3)=\frac{8!}{2!\cdot3!\cdot3!}\left(\frac{1}{6}\right)^8\approx3.334\cdot10^{-4}$$
where we do not have to take into account $X_2=X_4=X_5=0$ since $0!=1$ and $\frac{1}{6}^0=1$
R Simulation:
dice=c(1,2,3,4,5,6)
u = replicate(10^7,sample(dice,8,repl=T))
one=colSums(u==1)
two=colSums(u==2)
three=colSums(u==3)
four=colSums(u==4)
five=colSums(u==5)
six=colSums(u==6)
mean(three==2 & one==3 & six==3)
[1] 0.0003365
which agrees with our result fairly accurately.
Best Answer
Since you only ask why your intuition is wrong, that is the question I will answer.
When you add When you add $\frac{1}{6} + \frac{1}{6} + \frac{1}{6} $ you are double counting some of the possibilities. For example, the result $(6,6,6)$ was counted three times. And any outcome that has more than one $6$ was counted more than one time.