Asymptotic Value of $\sum_{n=1}^{N} \log(1+1/n) \log (\sum_{m=0}^{n} \binom{N}{m})$?

From Ian's answer, we have $$ \log \left( {\sum\limits_{m = 0}^n {\binom{N}{m}} } \right) = n\log \left( {\frac{N}{n}} \right) - (N - n)\log \left( {1-\frac{n}{{N}}} \right) + \mathcal{O}(\log N). $$ for $1 \leq n \leq N-1$. Now $$ \sum\limits_{n = 1}^{N-1} {\log \left( {1 + \frac{1}{n}} \right)\mathcal{O}(\log N)} = \mathcal{O}(\log N)\sum\limits_{n = 1}^{N-1} {\frac{1}{n}} = \mathcal{O}(\log ^2 N). $$ I showed in my answer to your other question (https://math.stackexchange.com/q/4099634) that $$ \sum\limits_{n = 1}^{N-1} {\log \left( {1 + \frac{1}{n}} \right)\left( {n\log \left( {\frac{N}{n}} \right) - (N - n)\log \left( {1 - \frac{n}{N}} \right)} \right)} = \frac{{\pi ^2 }}{6}N + \mathcal{O}(\log ^2 N). $$ Putting these together, shows that $$ \sum\limits_{n = 1}^N {\log \left( {1 + \frac{1}{n}} \right)\log \left( {\sum\limits_{m = 0}^n {\binom{N}{m}} } \right)} = \frac{{\pi ^2 }}{6}N + \mathcal{O}(\log ^2 N) $$ as $N\to +\infty$ (the term corresponding to $n=N$ on the left-hand side is $\mathcal{O}(1)$).

Not a complete answer, but maybe in the right direction.

One way to solve this is to look at $h(N+1) = f(N+1)-f(N)$, and then compute $\lim_{n \to \infty} f(N) =\sum_{N=1}^{\infty} h(N+1)$.

Here $h(N+1) = \log(1+1/(N+1)) \log \sum_{m=0}^{N+1} \binom{N+1}{n} + \sum_{n=1}^{N} \log(1+1/n)[ \log \sum_{m=0}^{n} \binom{N+1}{n} - \log \sum_{m=0}^{n} \binom{N}{n}]$.

The first term is $\log(1+\frac{1}{N+1}) \log 2^{N+1} = \log 2 \log((1+\frac{1}{N+1})^{N+1}) \to log 2 \cdot \log e = \log 2$ when $N$ is large.

I'm not so sure about the second term. When I compute it in Matlab it seems to be slightly above 1.

When I compute the original expression $f(N)/N$ into Matlab, I get a value around $\log 2 + 1$ which seems to support the conjecture that $\lim_{N \to \infty} h(N) = \log 2 + 1$ and $\lim_{N \to \infty} \frac{f(N)}{N}= \log 2 + 1$.