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.