Let $X$ be the event in which player $1$ wins the game, and $T$ denote the person whose turn it is to throw the die. We then find:
$$P(X | T = 1) = \frac{1}{6} + \frac{2}{6} P(X | T = 2) + \frac{3}{6} P(X | T = 3)$$
We know that:
$$P(X | T = 2) = \frac{3}{6} P(X | T = 1) + \frac{2}{6} P(X | T = 3)$$
$$P(X | T = 3) = \frac{2}{6} P(X | T = 1) + \frac{3}{6} P(X | T = 2)$$
$$= \frac{2}{6} P(X | T = 1) + \frac{3}{6} \left(\frac{3}{6} P(X | T = 1) + \frac{2}{6} P(X | T = 3)\right)$$
$$\iff P(X | T = 3) = \frac{21}{30} P(X | T = 1)$$
As such, it follows that:
$$P(X | T = 1) = \frac{1}{6} + \frac{2}{6} \left(\frac{3}{6} P(X | T = 1) + \frac{2}{6} P(X | T = 3)\right) + \frac{3}{6} P(X | T = 3)$$
$$= \frac{1}{6} + \frac{2}{6} \left(\frac{3}{6} P(X | T = 1) + \frac{2}{6} \frac{21}{30} P(X | T = 1)\right) + \frac{3}{6} \frac{21}{30} P(X | T = 1)$$
$$\iff P(X | T = 1) = \frac{30}{73} \approx 0.4110$$
This result can be confirmed using the following Python code:
from random import randint
s = [0, 0, 0]
n = 1000000
for _ in range(n):
p = 0
while True:
t = randint(1, 6)
if t == 6:
s[p % 3] += 1.0 / n
break
elif t == 2 or t == 4:
p += 2
else:
p += 1
print(s)
An example run returned:
[0.411106, 0.300798, 0.288096]
Solving a set of linear recurrences is indeed a good, elementary way to go, but if you solve the recurrences in the answer by @Canardini - which I did using wolfram alpha - you find that the answer is $E_X = 46656 = 6^6$. This is such a special number that you might wonder if there is a more fundamental explanation, and indeed there is, using more powerful theorems of Markov Chains.
Claim: If the desired string $x$ has the property that two copies of $x$ cannot overlap (which holds for $x = 123456$ in the OP question but does not hold for e.g. $x=111111$ or $x=121212$), then the expected time to first occurrence of $x$ is $6^L$ where $L$ is the length of $x$.
Consider a Markov Chain with $6^6$ states, where each state is a possible sequence in $\{1,2,3,4,5,6\}^6$ and records the last $6$ rolls. Each state can transition to $6$ states (i.e. it has "out-degree" $6$) with equal prob $1/6$. E.g. the state $\color{red}{1}13462$ can transition to $13462\color{blue}{j}$ where $\color{blue}{j}$ can be any of $\{1,2,3,4,5,6\}$. The red $\color{red}{1}$ represents the oldest die-roll result that has "aged out" and the blue $\color{blue}{j}$ represents the newest die-roll result. Note that each state also has "in-degree" $6$, i.e. only $6$ states can transition to it. (Self-loops are possible and count as both in-degree and out-degree.)
It is obvious such a Markov Chain is aperiodic, positive recurrent, irreducible, ergodic, etc., all the good stuff. Further, because every state's in-degree $=$ out-degree $= 6$, the chain's unique stationary distribution $\pi$ (also its limiting distribution) is the $6^6$-long vector whose every entry is $6^{-6}$.
A powerful (but somewhat "intuitively obvious?") theorem says that, if $\tau_{xx}$ is the revisit time from state $x$ back to state $x$, then:
Theorem: for a positive recurrent Markov Chain, with stationary distribution $\pi, E[\tau_{xx}] = 1 / \pi_x$ for any state $x$.
E.g. see Prop 2.6 of these notes or Theorem 1.19 of these notes or (for a slightly different version) wikipedia
IMHO this theorem is "intuitively obvious" in the following sense: the limiting distribution $\pi$ means in the long run the chain is going to spend $\pi_x$ fraction of the time in state $x$, so it only makes sense that the inter-visit time $\tau_{xx}$ has an expected value of $1/\pi_x$. However, such an "intuitive" argument is not rigorous, and the theorem has a non-trivial proof making use of positive recurrence.
Anyway, based on this theorem, and letting $x=123456$ the state we're interested in, we have $E[\tau_{xx}] = 1/6^{-6} = 6^6$. I.e., if we have just rolled $123456$, then the expected time to roll the next $123456$ is $6^6$. This isn't the same as the OP question. However, if we have just rolled $123456$, then none of these old roll-results can be part of the next $123456$, and therefore this is equivalent to rolling from the very beginning (when the "history" of rolls is the empty string). This is a direct result of the fact that two strings of $123456$ cannot overlap. So the same expected time $6^6$ also answers the OP question.
Addendum: for some other strings, this theorem also gives a quick way to find expected time of first occurrence. E.g. consider $y=111111$. The same theorem says that $E[\tau_{yy}] = 6^6$. But it is also obvious that revisit can either happen right away (if the next roll is $1$) or much later. I.e.:
$$E[\tau_{yy}] = 1 + (\frac16 \times 0 + \frac56 \times E[T_y])$$
where $T_y=$ time to first occurrence of $y$ starting with no useful history (including the case of starting from scratch, i.e. empty history). Solving for this we have:
$$E[T_y] = (6^6 - 1) \times \frac65 = 55986$$
which can be easily verified by solving the corresponding set of linear recurrences for the string $y=111111$.
Best Answer
You are assuming that the random variables $X_i$'s are independent but this is not the case.
Let $A_i$ be the event that $i$ occur, we want to compute $$P\left(\bigcap_{i=1}^6 A_i\right)=1-P\left( \bigcup_{i=1}^6 A_i^c\right)$$
We then compute $P\left( \bigcup_{i=1}^6 A_i^c\right)$ by inclusion-exclusion principle.
Notice that we have
$$P\left(\bigcap_{i=1}^m A_i^c\right)=\left( \frac{6-m}{6}\right)^{12}$$
By symmetry:
$$P\left( \bigcup_{i=1}^6 A_i^c\right)=\sum_{m=1}^6 (-1)^{m+1}\binom{6}{m}P\left(\bigcap_{i=1}^m A_i^c\right)$$