The question is arguably ambiguous -- do we consider side-length triples (2,3,4) and (2,4,3) to define the same triangle or two different ones? I shall begin with the latter version of the question because it's less fiddly, though I suspect the former is actually intended.
Three numbers are the sides of a triangle with positive area if and only if they
can be written as b+c,c+a,a+b with a,b,c positive.
These are the sides of an integer triangle with positive area if and only if
a,b,c are either all integers or all (odd) half-integers.
Hence $T_n$ equals
the number of triples of positive (a,b,c), either all integers or all halves of odd integers, with 2(a+b+c)=n.
Clearly
when n is even we must have a,b,c all positive integers with a+b+c=n/2; and when n is odd we must have a,b,c all positive integer - 1/2 -- say u-1/2,v-1/2,w-1/2 -- with u+v+w=(n+3)/2.
Now,
the number of ways to write $m$ as the sum of three positive integers equals the number of ways to write $m-3$ as the sum of three non-negative integers, which equals $\binom{m-1}{2}$
and therefore we want the first $n$ for which
either $n$ is even and $\binom{n/2-3}{2}>2016$ or $n$ is odd and $\binom{(n-3)/2}{2}>2016$.
Now, it happens that $2016=\binom{64}{2}$ exactly. So this first happens when
either $n$ is even and $n/2-3>64$, or $n$ is odd and $(n-3)/2>64$, whichever happens first; the former happens first at $n=136$ and the latter at $n=133$
and the answer is
n=133.
If we need to consider triangles the same when they have the same side lengths but in a different order, we now have to separate out the cases where they're all different (we only count 1/6 of these), where two are the same (we only count 1/3 of these), and where all three are equal (we count all of these). Note that
those conditions on the side lengths match exactly with identically-stated conditions on the parameters I've called a,b,c.
I am not going to do this right now because it's nearly 4am local time and I should have been in bed hours ago. Perhaps tomorrow, unless it turns out that the version I dealt with above is what was actually intended in which case I needn't bother.
[EDITED to add: Others have done this case so I won't bother. I'm sure they're right :-).]