My guess is you think that because your starting matrix is real, the diagonalization should be real as well. (Just a guess! If this isn't your issue with using the complex eigenvalues/vectors, do explain!)
This is not true, as your example already shows. Note that, as your Mathematica output gives, you'll want to diagonalize (up to order...) with $ D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -\frac{1}{2} - \frac{i\sqrt{3}}{2} & 0 \\ 0 & 0 & -\frac{1}{2} + \frac{i\sqrt{3}}{2} \end{pmatrix} $ and $ P = \begin{pmatrix} -\frac{1}{2} & 1 + i\sqrt{\frac{3}{2}} & 1 - i\sqrt{\frac{3}{2}} \\ -\frac{1}{2} & 1 - i\sqrt{\frac{3}{2}} & 1 + i\sqrt{\frac{3}{2}} \\ 1 & 1 & 1 \end{pmatrix} $ so that $ P^{-1} = \frac{1}{12}\begin{pmatrix} -4 & -4 & 8 \\ -i(2i + \sqrt{6}) & i(-2i + \sqrt{6}) & 2 \\ i(-2i + \sqrt{6}) & -i(2i + \sqrt{6}) & 2 \end{pmatrix} $ and $ P^{-1}AP = D $ or $ PDP^{-1} = A,$ whichever you prefer.
The reason for this is because the eigenvectors are complex conjugates, so even though the eigenvectors are complex, all of the complex parts will cancel out once you multiply everything through (recall, $(a+bi)(a-bi) = a^2 + b^2$!). Thus, as the other answer already suggests, the method for diagonalizing a real matrix is the same, regardless of what field (real or complex) the eigenvalues/vectors lie in. It just turns out that if you start with a real matrix, your eigenvectors, if complex, will always cancel out when you multiply everything through!