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
Total Number of elements is sample space =$6^3=216$.
Total number of elements in the sample space which contains no $5$ = $5^3=125$.
Thus total number of elements which having at least one $5$ is $=216-125=91$.
Update after OP request:
If there is only one $5$. Total Number of possibilities=$1×5×5×3C1=75$.
If there is only two $5$. Total Number of possibilities=$1×1×5×3C2=15$.
If there is only three $5$. Total Number of possibilities=$1×1×1=1$.
Total$=75+15+1=91$