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.
You are double counting the outcomes. You have to assume there is exactly one 2 instead of at least one 2, then exactly two 2's, and finally exactly three 2's:
Exactly one 2: $1\times 5\times 5\times \binom{3}{1}=75$
Exactly two 2's: $1\times 1\times 5\times \binom{3}2=15$
Exactly three 2's: $1\times 1\times1 =1$
This gives $75+15+1=91$ desirable outcomes.
An easier method would be to just find the number of outcomes with no 2's, then subtract that from the total number of outcomes:
$$6\cdot 6\cdot 6 - 5\cdot 5\cdot 5=216-125=91$$
Best Answer
Here is another approach that seems to work. Represent each die by a generating function (i.e. polynomial)
Since on each die there is only 1 outcome you want, a 5, there are five other outcomes you don't want. Call the outcomes you want y, and those you don't want x. The generating function for each die would then be $f(die)=5x+y$
Now multiply these functions to get the various outcomes:
$f(die1)*f(die2)*f(die3) = 125x^2 + 75x^2y + 15xy^2 + 1y^3$
You can read off the coefficients to see there are 125 permutations with no 5's rolled; 75 with just one 5 rolled (the coefficient to the y containing term), 15 with just two 5's rolled (the coefficient to the $y^2$ containing term) and only one roll with three 5's.
caveat my background in this area of math is very limited, so could be wrong that this approach can be generalized. I've no source for this approach except just noodling on it and tested it on simple cases.