As Stefan Kohl has pointed out it is not possible to give good lower bounds for $T(N)$ without making progress on the Collatz problem itself. However it is not difficult to understand what the truth is, and the argument below will yield a rigorous upper bound.

Instead of the map $C(n)$ as defined, it is better to consider $C_0(n) = n/2$ if $n$ is even, and $(3n+1)/2$ if $n$ is odd. For large $n$, this map multiplies by approximately $1/2$ or $3/2$ with probability $1/2$, and the first several iterations of the map behave independently. The crux of the Collatz problem is that this independent behavior roughly persists till the end making the iterates come down to zero -- that of course is hard, but it is easy to see that as many initial iterates as you like will behave independently just by looking at residue classes modulo a large power of $2$.

Now consider the random walk starting with a random real number $\alpha_0=\alpha$ chosen uniformly from $[0,1]$, and setting $\alpha_k = \alpha_{k-1} \times 3/2$ or $\times 1/2$ with equal probability. Then for each such realization take the max of $\alpha_k$ over all $k\ge 1$. In this setting you are asking for the probability that this maximum is at most $1/2$. (The half arises because of the difference between your $C(n)$ and the more natural $C_0(n)$.)

It is easy to upper bound the probability above by running the random walk for some number of steps. After $25$ steps, a computer calculation gave the upper bound $0.40641\ldots$, which agrees with Stefan Kohl's numerics.