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.
Alicia is playing the game. She wins if the sum total of values is at least $6$.
If the first toss results in a $4$ or a $5$, she won't get a chance to retoss, and she won't win.
Instant win: If she gets a $6$ on the first toss, then she has won quickly. This has probability $\dfrac{1}{6}$.
We now examine what happens if the first toss is a $1$, $2$, or $3$. Examine the cases one at a time.
First toss is a 1: If at first she gets a $1$, (probability $\frac{1}{6}$), then she gets to retoss. To win, she needs a $5$ or a $6$ on the second toss. The probability this sequence of events happens is $\dfrac{1}{6}\cdot \dfrac{2}{6}$.
First toss is a 2: In this case, to win she needs a $4$, $5$, or $6$ on the second toss. The probability this sequence of events happens is $\dfrac{1}{6}\cdot \dfrac{3}{6}$.
First toss is a 3: In this case she needs a $3$, $4$, $5$, or $6$ on the second toss. The probability this sequence of events happens is $\dfrac{1}{6}\cdot \dfrac{4}{6}$.
Thus the probabilility Alicia wins is $\dfrac{1}{6}+ \dfrac{1}{6}\cdot \dfrac{2}{6}+ \dfrac{1}{6}\cdot \dfrac{3}{6}+ \dfrac{1}{6}\cdot \dfrac{4}{6}$. Simplify.
Best Answer
Since the two die rolls are independent, it should come as no surprise that there are $6\cdot 6 = 36$ possible pairs of die rolls.
Of those rolls, only the following pairs have a sum of $7$: $$(1,6),\ (2,5),\ (3, 4),\ (4, 3),\ (5, 2),\ (6, 1)$$
Naturally, then, the likelihood of our two die rolls summing to $7$ would be: $$\frac 6{36} = \frac 16 \approx 16.6\%$$