We can prove a lower bound on the expression by considering only vertices immediately below a leaf:

$$\frac{[\sum_{v \in T \text{ not a leaf}}2^{h(v)}L(T(v))] + L(T)}{N(T)} \geq \frac{[\sum_{v \text{ below a leaf}}2^{h(v)}L(T(v))] + L(T)}{N(T)}.$$

For each such $v$, $L(T(v)) \geq 2$ and there are at most two leaves immediately above $v$, so

$$\frac{[\sum_{v \text{ below a leaf}}2^{h(v)}\cdot L(T(v))] + L(T)}{N(T)} \geq \frac{\frac{1}{2}[\sum_{v \text{ a leaf}}2^{h(v)-1}\cdot 2] + L(T)}{N(T)}.$$

Now $N(T) = 2L(T) - 1$, so we are working with an expression that is approximately

$$\frac{1}{4L(T)}\sum_{v \text{ a leaf}}2^{h(v)}.$$

This is (a quarter of) the average value for $2^{h(v)}$ when $v$ is a leaf.
To decrease the value without changing the height, one could attach a "V" on the top of some leaf $v'$ for which $3\cdot 2^{h(v')}$ is below the average.
One could also remove a "V" with two leaves that produce an above-average value, so long as at least one leaf of height $h$ remains.

So the minimum average for some fixed height $h$ will be obtained by a tree that is (approximately) a full binary tree of height $h'$ with a single branch growing up to height $h$.
Then

\begin{eqnarray}
\frac{1}{4L(T)}\sum_{v \text{ a leaf}}2^{h(v)} & \approx & \frac{1}{4(h - h' + 2^{h'})} \left[\left(\sum_{i = h'}^{h} 2^i\right) + \left(2^{h'}\right)^2 \right]\\
&\geq& \frac{1}{4}\frac{1}{(h - h' + 2^{h'})} \left[2^h + 2^{2h'}\right]\\
& \geq & \frac{1}{4}2^{h/2}
\end{eqnarray}
for all sufficiently large $h$ and all $h' < h$. Indeed, for small $h'$, $\frac{1}{h}2^{h}$ dominates; if $h'$ is large enough for $2^{h'}$ to dominate the denominator and $\frac{1}{2^{h'}}2^h < 2^{h/2}$, then $h' > h/2$ and $\frac{1}{2^{h'}}2^{2h'} > 2^{h/2}$.