Consider the spacetime bivectors $\gamma_{12}$ and $\gamma_{30}$ with the metric $\eta=(1, -1, -1, -1)$. I would calculate their dot product to be
$\gamma_{12} \cdot \gamma_{30} = \frac{1}{2}\left(\gamma_{12} \gamma_{30} + \gamma_{30} \gamma_{12}\right) = \frac{1}{2}\left(\gamma_1 \wedge \gamma_2 \wedge \gamma_3 \wedge \gamma_0 + \gamma_3 \wedge \gamma_0 \wedge \gamma_1 \wedge \gamma_2 \right) = \frac{-I_{0123} - I_{0123}}{2} = -I_{0123}$, where $I_{0123}$ is the pseudoscalar.
However, as the two bivectors should be orthogonal I would have expected the result to be zero, which is the case for bivector pairs sharing an index like $\gamma_{12}$ and $\gamma_{23}$. Is my calculation wrong or what is wrong with my reasoning?
This is very incorrect. First, this dot product should yield a scalar, not a pseudoscalar. Second, you're probably getting the formula $$ a\cdot b = \frac12(ab + ba) $$ from the specific case that $a$ and $b$ are vectors.
If $a_1,\dotsc,a_k$ and $b_1,\dotsc,b_k$ are vectors, then for two $k$-blades we can find that $$ (a_k\wedge a_{k-1}\wedge\dotsb\wedge a_1)\cdot(b_1\wedge b_2\wedge\dotsb\wedge b_k) = \det\bigl(a_i\cdot b_j\bigr)_{i,j=1}^k. $$ So when $k=2$ $$ (a_1\wedge a_2)\cdot(b_1\wedge b_2) = (a_2\cdot b_1)(a_1\cdot b_2) - (a_1\cdot b_1)(a_2\cdot b_2). $$
In your particular case though its even simpler. For $\gamma_{12}\cdot\gamma_{30}$ the result is supposed to be a scalar, and in fact $$ \gamma_{12}\cdot\gamma_{30} = \langle\gamma_1\gamma_2\gamma_3\gamma_0\rangle. $$ But clearly $\gamma_1\gamma_2\gamma_3\gamma_0$ is a pseudoscalar and has no scalar part, so $\gamma_{12}\cdot\gamma_{30} = 0$.
I can't say much more without knowing your particular definition of $\cdot$ or what you're learning from.