The answer to this question depends on what is a "different" outcome.
Interpretation 1: The possible outcomes are the multisets $\{a,b,c\}$ such that $a,b,c \in \{1,2,\ldots,6\}$. This is justified since dice are identical objects, and so e.g. the outcome $\{1,1,2\}$ is counted as the same thing as $\{1,2,1\}$ and $\{2,1,1\}$.
In general, using a stars-and-bars argument, the number of such multisets is ${n+5} \choose 5$. For example, $\{1,4,4,6\}$ is counted by $\underbrace{\star}_{\text{one } 1} | \; | \; | \underbrace{\star \star}_{\text{two } 4\text{'s}} | \; | \underbrace{\star}_{\text{one } 6}$ (i.e. we generate a string with $n$ stars and $5$ bars, and the position of the bars determines the number of copies of each element in the multiset).
In GAP, this is implemented as NrUnorderedTuples([1,2,3,4,5,6],3);. It is also described by Sloane's A000389.
Interpretation 2: The possible outcomes are the sums $a+b+c$ of the multisets $\{a,b,c\}$ such that $a,b,c \in \{1,2,\ldots,6\}$. In general, anything from $n$ to $6n$ is a possible outcome. Hence the number of possible outcomes is $5n+1$.
You also have correctly surmised that rolling $n$ dice and rolling the same die $n$ times are equivalent.
** How many outcomes are in the event where nobody rolls a six?
If they can't roll a six, there are 5 other numbers to roll, and either coin-flip is still allowed. So each person has $2\times 5=10$ possible outcomes. Since there are 5 people, there are $10^5$ possible outcomes.
** How many outcomes are in the event where at least one person rolls a six?
Well, either there are no sixes rolled or at least one person rolls a six, so this result is the difference of your answer and the answer above $12^5 - 10^5$.
Best Answer
In general , we think that dice are different , so there are $6^n$ different possible outcomes when dice are distinct .
However, when the dice are identical , we only consider that how many different number will appear ,we can do it by stars and bars such that
Assume that the gaps between the bars means numbers on die . For example $$1|2|3|4|5 |6 $$ .Now , if we have $n$ stars , how many ways are there to see possible out comes ?
Answer is $$C(n+6-1, 5)$$ . Because ,we think the arrangements of $n$ identical stars and $5$ identical bars. You can think it like combination with repetition.
For example , if $n=4$ then $*|**|||*|$ means $(1,2,2,5)$ , $|||||****$ means $(6,6,6,6)$
If $n=3$ ,then $||*||*|*$ means $(3,5,6)$