Inner product structure on geometric algebra?

Yes. The natural inner product is

$$A\star B=\langle A^\sim B\rangle_0=\langle A\,B^\sim\rangle_0=\langle B^\sim A\rangle_0=\langle B\,A^\sim\rangle_0$$

$$=\langle A^\sim\bullet B\rangle_0=\langle A^\sim\,\lrcorner\,B\rangle_0=\langle A^\sim\,\llcorner\, B\rangle_0,$$

the scalar part of the geometric product (or any of these four products), with one factor reversed. This is analogous to the Frobenius inner product of matrices, $\text{tr}(A^TB)$. It's the familiar sum of products of corresponding components, with respect to a standard orthonormal basis $\{1,e_1,e_2,e_1e_2,e_3,e_1e_3,e_2e_3,e_1e_2e_3,\cdots\}$. For example in 2D,

$$(\alpha+\alpha_1e_1+\alpha_2e_2+\alpha_{12}e_1e_2)\star(\beta+\beta_1e_1+\beta_2e_2+\beta_{12}e_1e_2)=\alpha\beta+\alpha_1\beta_1+\alpha_2\beta_2+\alpha_{12}\beta_{12}.$$

Terms of different grade are orthogonal. Terms of the same grade have inner product

$$(a_1\wedge a_2\wedge\cdots\wedge a_k)\star(b_1\wedge b_2\wedge\cdots\wedge b_k)=(a_k\wedge\cdots\wedge a_2\wedge a_1)\bullet(b_1\wedge b_2\wedge\cdots\wedge b_k)$$

$$=\det[a_i\cdot b_j].$$

(I'm assuming that the underlying vector space has an inner product; then $\star$ is also an inner product. But this $\star$ extends any symmetric bilinear form on the vector space to the whole algebra, regardless of positive-definiteness or non-degeneracy.)