For $s>0$ one has $ \psi^{(s)}(\frac{1}{2}) = s! \cdot \zeta(s+1, \frac{1}{2}) \cdot (-1)^{s+1} $.
E.g.
$ \psi^{(1)}(\frac{1}{2}) = 3 \cdot \zeta(2) $
$ \psi^{(2)}(\frac{1}{2}) = -14 \cdot \zeta(3) $
$ \psi^{(3)}(\frac{1}{2}) = 90 \cdot \zeta(4) $
etc.
However, Mathematica gives
$ \psi^{(-1)}(\frac{1}{2}) = \frac{\ln(\pi)}{2} $
$ \psi^{(-2)}(\frac{1}{2}) = \ln(A^\frac{3}{2}2^\frac{5}{24}\pi^\frac{1}{4}), $ where $A$ is the Glaisher-Kinkelin Constant
$ \psi^{(-3)}(\frac{1}{2}) = \ln(A^\frac{1}{2}2^\frac{1}{16}\pi^\frac{1}{16}) + \frac{7 \zeta(3)}{32\pi^2} $
and I'm not sure how these are obtained since negative integers plugged into the formula above yields wonky results.
First off we have for reference this formula which states for all $n\in\Bbb N$ $$ \psi ^{(n)}\left(\tfrac{1}{2}\right)=(-1)^{n+1} \left(2^{n+1}-1\right) n! \zeta (n+1). $$ Note that this is not the same formula as given in your question statement. As for negative values, I was not able to find a closed-form for all $-n\in\Bbb N$ at $z=1/2$ but there is one for $-2n$ with $n\in\Bbb N$ and $z=1/2$ given here. Furthermore, we do have a formula for general $-n$ and $z$ given here which can be evaluated at $z=1/2$. With a little extra work you could further simplify this to the special cases you obtained using Mathematica using the well-known properties of the polygamma function.