You have the total number of solutions. Let's first count the number of solutions in which we do not have all three of $a$, $b$, $c$ pairwise distinct. We'll take them out directly before figuring out our overcount.
How many solutions are there with at least two of the numbers equal? Say $a=b=k$. Then we must have $2k\leq n$, or $k\leq \lfloor\frac{n}{2}\rfloor$. Once you have that equal value, the third variable has only one possibility.
So you have $\lfloor\frac{n}{2}\rfloor+1$ such solutions (to include $0$); then you get to choose the "odd man out" in $3$ different ways. So we have $3\lfloor\frac{n}{2}\rfloor+3$ possibilities. However, you have overcounted a solution with $a=b=c$ if there is one; there is one if and only if $n\equiv 0\pmod{3}$. So the number of solutions with at least two of the numbers equal is $\begin{align*} 3\left\lfloor\frac{n}{2}\right\rfloor+1&\qquad\text{if }n\equiv 0\pmod{3}\\ 3\left\lfloor\frac{n}{2}\right\rfloor+3 &\qquad\text{otherwise.} \end{align*}$
For example, take $n=11$. The solutions with duplicates have $(0,0,11)$, $(1,1,9)$, $(2,2,7)$, $(3,3,5)$, $(4,4,3)$, and $(5,5,1)$, and each of them can be assigned to $a$, $b$, and $c$ in three different ways. This gives 18 solutions; $\lfloor\frac{n}{2}\rfloor = \lfloor\frac{11}{2}\rfloor = 5$, and since $11$ is not divisible by $3$, the formula gives $3(5)+3 = 18$.
Now say $n=15$; according to our formula, we should get $3\lfloor\frac{15}{2}\rfloor + 1 = 22$ solutions. What are they? There are three solutions for each of the following multiset of values: $(0,0,15)$, $(1,1,13)$, $(2,2,11)$, $(3,3,9)$, $(4,4,7)$, $(6,6,3)$, and $(7,7,1)$; and then there is one solution with $a=b=c=5$. Total: 22 solutions.
Okay, so now we know how many solutions have at least one equal sign. If we take them out of the total solutions, then we are left with only solutions in which $a$, $b$, and $c$ are all distinct. For each set of values that yield $n$, there are $3!=6$ ways of "distributing" the values among $a$, $b$, and $c$. So if we divide the number we have by $6$, we will get the answer you seek.
In summary, we get: $\begin{align*} \frac{1}{6}\left(\binom{n+2}{2} - 3\left\lfloor\frac{n}{2}\right\rfloor -1\right) &\qquad\text{if }n\equiv 0\pmod{3}\\ \frac{1}{6}\left(\binom{n+2}{2} - 3\left\lfloor\frac{n}{2}\right\rfloor -3\right) &\qquad\text{otherwise.} \end{align*}$
Let's test the formula. If $n=11$, then the solutions with $a\gt b \gt c\geq 0$ are: $(10,1,0)$, $(9,2,0)$, $(8,3,0)$, $(7,4,0)$, $(6,5,0)$, $(8,2,1)$, $(7,3,1)$, $(6,4,1)$, $(6,3,2)$, and $(5,4,2)$. That's it, because if $c=3$, then we would neeed $b\geq 4$ and $a\geq 5$, and that's already too much. So we get a total of ten solutions. According to the formula we found above, we should have $\frac{1}{6}\left(\binom{13}{2} - 3\left\lfloor\frac{11}{2}\right\rfloor -3\right) = \frac{1}{6}(78 - 15-3)=\frac{60}{6} = 10.$ Good.
Now, if $n=15$, the solutions with $a\gt b\gt c\geq 0$ are: $(14,1,0)$, $(13,2,0)$, $(12,3,0)$, $(11,4,0)$, $(10,5,0)$, $(9,6,0)$, $(8,7,0)$, $(12,2,1)$, $(11,3,1)$, $(10,4,1)$, $(9,5,1)$, $(8,6,1)$, $(10,3,2)$, $(9,4,2)$, $(8,5,2)$, $(7,6,2)$, $(8,4,3)$, $(7,5,3)$, and $(6,5,4)$; a total of 19 solutions. According to the formula, we have: $\frac{1}{6}\left(\binom{17}{2}-3\left\lfloor\frac{15}{2}\right\rfloor -1\right) = \frac{1}{6}(136 - 21 -1) = \frac{114}{6} = 19.$
Well, looks right, at any rate...
To count the number of solutions with $a\geq b\geq c$, we can take the total above for those with $a\gt b\gt c$. Then we add one for each solution with at least two numbers equal, namely $\lfloor\frac{n}{2}\rfloor+1$ solutions.