Proof $\sum\limits_{k=1}^n \binom{n}{k}(-1)^k \log k = \log \log n + \gamma +\frac{\gamma}{\log n} +O\left(\frac1{\log^2 n}\right)$

Thanks go to Andrew, J. M., and David Speyer! The following solution leans heavily on what these three have already posted.

(In the interest of having a complete solution I've added the argument that gets from the OP's sum to Andrew's reformulation of it.)

Since $$\int_0^1\sum_{m=1}^{k-1}\frac1{x+m}\, dx = \sum_{m=1}^{k-1}(\log(m+1) - \log m) = \log k,$$ we can rewrite the original formula as $$\sum_{k=1}^n \binom{n}{k}(-1)^k \log k= \sum_{k=2}^n \binom{n}{k}(-1)^k\int_0^1\sum_{m=1}^{k-1}\frac1{x+m}\, dx = \int_0^1 \sum_{m=1}^{n-1} \frac1{x+m} \sum_{k=m+1}^n \binom{n}{k}(-1)^k\, dx.$$ Since alternating row sums of binomial coefficients are easy to evaluate (see, for instance, Concrete Mathematics, Identity 5.16), this becomes (and then switching the index back to $k$) $$\int_0^1 \sum_{m=1}^{n-1} \frac1{x+m} (-1)^{m-1} \binom{n-1}{m}\, dx = \int_0^1 \sum _{k=1}^{n-1}\binom{n-1}{k} \frac{(-1)^{k-1}}{k+x} \, dx$$ $$ = \int_0^1 \left(\frac{1}{x} - \sum _{k=0}^{n-1}\binom{n-1}{k} \frac{(-1)^k}{k+x}\right) \, dx.$$ The remaining binomial sum is actually the partial fractions decomposition of $\frac{(n-1)!}{x(x+1)\cdots (x+n-1)}$. (This is identity 5.41 in Concrete Mathematics. From another perspective - also discussed in Concrete Mathematics - the binomial sum is $(-1)^n$ times the $n-1$ difference of $\frac{1}{x} = (x-1)^{\underline{-1}}$. Applying the finite difference rule $\Delta x^{\underline{m}} = m x^{\underline{m-1}}$ successively $n-1$ times thus gets us to $\frac{(n-1)!}{x(x+1)\cdots (x+n-1)}$.)

Thus our original sum is equivalent to $$\int_0^1 \left(\frac{1}{x} - \frac{(n-1)!}{x(x+1)\cdots (x+n-1)}\right) dx = \int_0^1 \left(\frac{1}{x} - \frac{\Gamma(n) \Gamma(x)}{\Gamma(n+x)}\right) dx,$$ which is the formula Andrew mentions in the comments.

Part 2: Rewriting the expression.

Now, rewrite like so: $$\int_0^1 \left( \frac{1}{x} - \frac{\Gamma(n) \Gamma(x)}{\Gamma(n+x)} \right) dx = \int_0^1 \left( \frac{1}{x}\left(1- \frac{\Gamma(n) \Gamma(x+1)}{\Gamma(n+x)} \right)\right) dx.$$

For $0 < x < 1$, $$\Gamma(x+1) = 1 - \gamma x + \frac{\zeta(2) + \gamma^2}{2}x^2 + O(x^3).$$ (This is the Maclaurin series for $\Gamma(x+1)$; see Expression 8.321 in Gradshteyn and Ryzhik. The only reason I know this is because I had to track it down for a paper I wrote a couple of years ago.) Also, $$\frac{\Gamma(n)}{\Gamma(n+x)} = n^{-x}\left(1 + O\left(\frac{1}{n}\right)\right).$$ (See the DLMF.)

Putting all this together means we want the asymptotic value of \begin{equation} \int_0^1 \frac{1-n^{-x}\left(1+ O\left(\frac{1}{n}\right)\right)\left(1 - \gamma x + \frac{\zeta(2) + \gamma^2}{2}x^2 + O(x^3)\right)}{x} dx. \tag{1} \end{equation}

Part 3: Obtaining the dominant terms.

Following David Speyer and J.M., we'll first extract what turns out to be the dominant part of (1): $$\int_0^1 \frac{1-n^{-x}}{x}dx = \int_0^1 \frac{1-e^{-x \log n}}{x}dx = \int_0^{\log n} \frac{1-e^{-u}}{u} du = \text{Ein}(\log n),$$ where $\text{Ein}(x)$ is the complementary exponential integral. Now, $\text{Ein}(x) = E_1(x) + \log x + \gamma$, where $E_1(x)$ is the usual exponential integral (again, see the DLMF), and $E_1(x) < e^{-x} \log (1 + 1/x)$ (DLMF once again), so putting all of this together we have $$\int_0^1 \frac{(1-n^{-x})}{x}dx = \log \log n + \gamma + O\left(\frac{1}{n}\right).$$

Part 4: Obtaining the remaining terms.

Now we consider the rest of (1). This is $$\left(1+ O\left(\frac{1}{n}\right)\right)\int_0^1 n^{-x}\left(\gamma - \frac{\zeta(2) + \gamma^2}{2}x + O(x^2)\right) dx$$ $$=\left(1+ O\left(\frac{1}{n}\right)\right)\left(\gamma \left(\frac{n-1}{n \log n}\right) - \frac{\zeta(2) + \gamma^2}{2}\left(\frac{n - \log n -1}{n (\log n)^2}\right) + O\left(\frac{1}{(\log n)^3}\right)\right)$$ $$=\frac{\gamma }{\log n} - \frac{\zeta(2) + \gamma^2}{2(\log n)^2} + O\left(\frac{1}{(\log n)^3}\right),$$ which is the rest of the expression requested by the OP.

Here is a non-rigorous approach which gives the right leading order term first two terms. I'm writing it here in the hope that someone will follow up and turn it into an actual proof. This post is CW, in case some one wants to add to what I've come up with.

I start with Andrew's formula: $$\sum_{k=1}^n (-1)^k \binom{n}{k} \log k = \int_0^1 \left( \frac{1}{x} - \frac{\Gamma(n) \Gamma(x)}{\Gamma(n+x)} \right) dx = \int_0^1 \left( \frac{1}{x} - \frac{(n-1)!}{x(x+1)(x+2) \cdots (x+n-1)} \right) dx$$ $$=\int_0^1 \frac{dx}{x} \left( 1- \frac{1}{(1+x)(1+x/2) \cdots (1+x/(n-1))} \right)$$

Let's look at that denominator. $$\prod_{k=1}^{n-1} (1+x/k) = \exp \left( \sum_{k=1}^{n-1} \log \left(1+\frac{x}{k}\right) \right) \approx \exp \left( \sum_{k=1}^{n-1} \frac{x}{k} \right) \approx e^{x \log n}.$$

I could make these estimates more precise, but I'm not going to bother because I don't have a rigorous proof. So, roughly, we want to compute $$\int_0^1 \frac{1-e^{-x \log n}}{x} dx = \int_0^{\log n} \frac{1-e^{-u}}{u} du$$ where we have substituted $u = x \log n$. Let $[u>1]$ be $1$ for $u>1$ and $0$ for $u<1$. Then we can write this integral as $$\int_0^{\log n} \frac{[u>1] du}{u} + \int_0^{\log n} \frac{1-[u>1]-e^{-u}}{u} du = \log \log n + \int_0^{\log n} \frac{1-[u>1]-e^{-u}}{u} du$$

One can check that the integral $\int_0^{\infty} (1-[u>1]-e^{-u})/u \ du$ converges to some constant $\gamma$, as JM points out below. So, if we make our estimates precise, this method will give $$\log \log n + \gamma + o(1) \quad \mbox{as $n \to \infty$}.$$

"Concrete" methods should be able to improve the approximation $\sum_{k=1}^{n-1} \log \left(1+\frac{x}{k}\right) \approx x \log n$ a good deal. Once we see what that improved version looks like, we can try to figure out what to do with the rest of the argument.