Although $\eta(1)$ is known to be $\ln(2)$, I have not seen an analytically calculated value for $\eta(\frac{1}{2});$
$$\eta\left(\frac{1}{2}\right) = \sum_{n=1}^{\infty}\frac{(-1)^{(n+1)}}{\sqrt{n}}$$
A web calculator gives the value to be 0.6, which seems to be right.
On
A careful computation shows that the numerical value is $$0.6048986434216303702472...$$ which is not $0.6$. One should be aware that the above series converge really slowly.
As Claude Leibovici indicates, one can relate its value to the Riemann's Zeta function value at $1/2$. However, as far as I know, there is no analytic formula of $\zeta(1/2)$, so this is why you haven't seen an "analytically calculated value for $\eta(1/2)$".
EDIT2: As pointed again in the comments by leftaroundabout, I missread the OEIS link given in the answer of R. J. Mathar. What equals $$\gamma/2 + \pi/4 - (1/2 + \sqrt{2})\log(2) + \log(\pi)/2,$$where $\gamma$ is the Euler-Mascheroni constant, is $\eta'(1/2)/\eta(1/2)$ not $\eta(1/2)$.
Isn't just $$\eta\left(\frac{1}{2}\right)=\sum_{n=1}^\infty\frac{(-1)^{(n+1)}}{\sqrt{n}}=\left(1-\sqrt{2}\right) \zeta \left(\frac{1}{2}\right)\approx 0.6048986434$$
Edit
Remember the general relation $$\eta\left(s\right)=\left(1-2^{1-s}\right) \zeta (s)$$ If you want a quick and dirty shortcut evaluation, for $0 \leq s \leq 1$, you could use $$\eta\left(s\right)=\frac 12+\left( \log (2)-\frac{1}{2}\right)\, s^{0.895}$$