The integral form of the remainder is:
$$R_n(x) = \int_a^x \frac{f^{(n+1)}(t)}{n!}(x-t)^n\;dt$$
this is exact, not an approximation, not a bound. If we now bound $|f^{(n+1)}(t)|$ on $[a,x]$ by some $M$ then
\begin{align}|R_n(x)|
&= \left|\int_a^x \frac{f^{(n+1)}(t)}{n!}(x-t)^n\;dt\right|
= \frac{1}{n!}\left|\int_a^x f^{(n+1)}(t)(x-t)^n\;dt\right|\\
&\overset{(*)}{\leq} \frac{1}{n!}\left|\int_a^x M|x-t|^n\;dt\right| \\
&= \frac{M}{n!}\frac{|x-a|^{n+1}}{n+1}
= \frac{M}{(n+1)!}|x-a|^{n+1}
\end{align}
This is of the form of the Taylor inequality. Taking the smallest such $M$ we get the typical inequality, $|R_n(x)| \leq U_n(x) \equiv \frac{\max_{t\in[x,a]}|f^{(n+1)}(t)|}{(n+1)!} |x-a|^{n+1}$.
My reason for starring $(*)$ is to emphasize the bluntness of this approximation. If $M|x-t|^n$ is a sufficiently loose upperbound for $|f^{(n+1)}(t)(x-t)^n|$ then the bound on the remainder could be way off.
In the specific case of $f(x) = \ln(1+x)$ we have $f^{(n+1)}(x) = (-1)^n \frac{n!}{(1+x)^{n+1}}$. As you correctly noted $|f^{(n+1)}(x)| \leq 2^{n+1} n!$ on $[-\tfrac12,0]$. The true remainder is:
\begin{align}
R_n(-\tfrac12) &= \frac{1}{n!}\int_0^{-\tfrac12} f^{(n+1)}(t)(-\tfrac{1}{2} - t)^n\;dt
= \frac{(-1)^n}{n!} \int_0^{-\tfrac12} (-1)^n\frac{n!}{(1+t)^{n+1}} (\tfrac12 + t)^n \;dt\\
&= \int_0^{-\tfrac12} \frac{1}{1+t} \left(\frac{\tfrac12 + t}{1+t}\right)^n \;dt = \frac{1}{2^n}\int_0^{-\tfrac12} \frac{1}{1+t} \underbrace{\left(2 - \frac{1}{1+t}\right)^n}_{(**)} \;dt
\end{align}
Using wolfram alpha to compute this integral for specific $n$ we can find $|R_{11}(-\tfrac12)| \approx 3.79\times 10^{-5}$ and $|R_{12}(-\tfrac12)| \approx 1.76\times 10^{-5}$ and $|R_{13}(-\tfrac12)| \approx 8.2\times 10^{-6}$. So the minimum $n$ is actually $13$. Indeed these match up with a direct calculation.
On the other hand $U_n(x) = \frac{2^{n+1}}{n+1} |x|^{n+1}$. This gives, as you found, $U_n(-\tfrac12) = \frac{1}{n+1}$ which is indeed a pretty dreadful upperbound on $|R_n(-\tfrac12)|$. In fact even to just get $U_n(-\tfrac12) < 10^{-1}$ we need $n>9$ terms, but at that point the true remainder is already $|R_9(-\tfrac12)|<10^{-3}$.
Note that the poorness of the approximation is due to $(**)$ in the integrand. For large $n$ it is only near $1$ very close to $0^-$ and is basically $0$ everywhere else on $[-\tfrac12,0]$. In other words the only significant contribution of $\frac{1}{1+t}$ to the integral is restricted to a small neighborhood near $0^-$. It should therefore be obvious that using a bound of the $n+1$st order derivative, far away from contributing portions of the integrand, could lead to a bad bound $U_n(x)$.
A bit beyond
With a little bit more work we can get a much better asymptotic bound on the true remainder. Starting from the $(**)$ step and substituting $y = \frac{1}{1+t} - 1$,
$$R_n(-\tfrac12)
= \frac{1}{2^n} \int_0^1 \underbrace{(y+1)}_{\frac{1}{1+t}} \cdot\underbrace{(1-y)^n}_{ \left(2 - \frac{1}{1+t}\right)^n} \cdot \underbrace{-\frac{dy}{(1+y)^2}}_{dt}
= -\frac{1}{2^n}\int_0^1 \frac{(1-y)^n}{1+y}\;dy
$$
Note that the integrand is positive in $(0,1)$, so the integral is positive, hence $R_n(-\tfrac12) < 0$. This makes sense as the terms of the Taylor approximations $-\sum_{n=1}^N \frac{1}{n\cdot 2^n}$ is decreasing in $N$, so it converges to $\ln(0.5)$ from above.
Using integration by parts with $u = \frac{1}{1+y}$ and $dv = (1-y)^n\;dy$ we get
\begin{align}
|R_n(-\tfrac12)|
&= \frac{1}{2^n}\left( \bigg[\underbrace{\frac{1}{1+y}}_{u}\cdot \underbrace{- \frac{(1-y)^{n+1}}{n+1}}_{v}\bigg]_0^1 - \int_0^1 \underbrace{\frac{(1-y)^{n+1}}{n+1}}_{v} \cdot \underbrace{-\frac{dy}{(1+y)^2}}_{du} \right)\\
&= \frac{1}{2^n} \frac{1}{n+1} - \frac{1}{2^n}\frac{1}{n+1}\int_0^1 \frac{(1-y)^{n+1}}{(1+y)^2} \;dy \\
&\leq \frac{1}{2^n(n+1)} - \frac{1}{2^n(n+1)}\int_0^1 \frac{(1-y)^{n+1}}{(1+(1))^2} \;dy \\
&= \frac{1}{2^n(n+1)} - \frac{1}{2^n(n+1)}\frac{1}{4} \frac{1}{n+2} \\
&= \frac{1}{2^n(n+1)} - \frac{1}{2^{n+2}(n+1)(n+2)} \\
\end{align}
To see just how much better this is than Taylor's approximation for $\ln(0.5)$, consider $n=100$. Taylor's gives the bound $U_{100}(-\tfrac12) = \frac{1}{100+1} \approx 10^{-2}$, whereas using the first term above gives us
$$
|R_{100}(-\tfrac12)|
\leq \frac{1}{2^{100}\cdot (100 + 1)}
= \frac {1}{\left(2^{10}\right)^{10} \cdot 101}
\leq \frac{1}{\left(10^3\right)^{10} \cdot 10^2}
= 10^{-32}
$$
To summarize, $$R_n(-\tfrac12) = -\frac{1}{2^n(n+1)} + \Theta\left(\frac{1}{2^n \cdot n^2}\right)$$
Best Answer
Not an answer, just a trick to expand the tangent.
By successive differentiations,
$$y(x)\cos x=\sin x$$
$$y'(x)\cos x-y(x)\sin x=\cos x$$
$$y''(x)\cos x-2y'(x)\sin x-y(x)\cos x=-\sin x$$
$$y'''(x)\cos x-3y''(x)\sin x-3y'(x)\cos x+y(x)\sin x=-\cos x$$
$$\cdots$$
Now with $x=\dfrac\pi4$, we have $\sin x=\cos x$ and the relations simplify to
$$y = 1$$
$$y' -y = 1$$
$$y'' -2y' -y =- 1$$
$$y''' -3y'' -3y' +y =- 1$$
$$\cdots$$