cQuents*, 10 bytes

O920A$/6^$

That * there is because, while this should be a specification-correct program that does the correct computation, the only interpreter available does not implement importing any OEIS sequence, but just a few. I suppose this might suggest this is not a valid submission, in which case I would still want to publish it but mark it as non-competing, which I don't think the leaderboard below the question understands as a concept

Anyways, this code does this:

• O920A$ the input-th item of the OEIS sequence A000920
• / divided by
• 6^$ 6 to the power of the input

Dyalog APL, 21 bytes

{≢⍸6=(≢∪)¨⍳⍵/6}÷(6∘*)

Bruteforce solution :)

• ⍳⍵/6 n-dimensional array of all possibilities of rolling $n$ dice, each element is a vector of the dice values
• ≢ count the elements
• ⍸ where it is true that
• (≢∪)¨ the length of the unique elements
• 6= is equal to 6
• ÷ and divide by
• 6∘* 6 to the power of $n$

Thunno 2, 14 bytes

6RẉDæḳ6R⁼;,ḷẸ\

Attempt This Online!

Note: brute-force, so times out for $n\ge5$ on ATO, but I have verified it with $n=6$ on my computer.

Explanation

6RẉDæḳ6R⁼;,ḷẸ\  # Implicit input
6R              # Push [1..6]
  ẉ             # Cartesian power with input
   D            # Duplicate the list
    æ    ;      # Filter by:
     ḳ          #  Sorted uniquify
      6R⁼       #  Exactly equal to [1..6]?
          ,     # Pair with original list
           ḷ    # Length of each list
            Ẹ   # Dump
             \  # Swapped divide
                # Implicit output

Dyalog APL, 19 bytes

(with index origin zero)

{-/(!∘6×⍵*⍨6÷⍨⊢)⍳7}

This isn't bruteforce! The number of cases where all six faces appear are the OEIS sequence A000920.

\[ P_n = \frac{\mathsf{OEIS}_{\text{A000920}}(n)}{6^n} \]

which is

\[ \frac {6!} {6^n} \left\lbrace {n \atop 6}\right\rbrace \]

where $\left\lbrace{n\atop k}\right\rbrace$ are the Stirling numbers of the second kind, which expand like this:

\[ \frac 1 {6^n} \sum_{i=0}^6 \left(-1\right)^{6-i} i^n \binom 6 i \]

(where $\binom n k$ are binomial coefficients) which simplifies to

\[ \sum_{i=0}^6 \left(-1\right)^i \binom 6 i\left(\frac i 6\right)^n \]

Note that the $i=0$ term is zero, but it is useful for me because it means the first term of the alternate sum has a positive coefficient.

Code explanation:

• ⍳7 To the list of numbers 0 through 6, apply the following function:
  • !∘6 $\binom 6 x$
  • × times
  • 6÷⍨⊢ $\frac x 6$
  • ⍵*⍨ to the power of $n$
• -/ and then take the alternate sum

Kinda sad to see a bruteforce solution is smaller than the closed form, I'll try and find a language where this is even shorter :)

Side note: this seems to have some floating point error which means the results for $n \in \left\{ 3, 4, 5 \right\}$ aren't exactly zero but in the range of $10^{-16}$ with ⎕fr←647 and $10^{-34}$ with ⎕fr←1287. either of these are both within the 6 decimal places of precision required, but just in case you cared about the pure computation, I think this is just error stacking up.