While this is a different measure than the two you talk about, a typical measure of "orthogonality" for lattice basis reduction is the orthognality defect.
The motivation in defining it is what is known as Hadamard's Inequality.
Let $\mathbf{B}$ be a matrix with columns $\mathbf{b}_1,\dots,\mathbf{b}_n$. Then $\det \mathbf{B}\leq \prod_{i=1}^n \lVert \mathbf{b}_i\rVert_2$, with equality if and only if $\mathbf{b}_i$ are mutually orthogonal.
Recall thatthe determinant is invariant under orthogonal transformations.
This suggests that we can freely switch $\det \mathbf{B} = \det \widehat{\mathbf{B}}$, where $\widehat{\mathbf{B}}$ is the Gram-Schmidt orthogonalization of $\mathbf{B}$.
Anyway, this suggests that the ratio
$$\frac{\prod_{i = 1}^n \lVert \mathbf{b}_i\rVert_2}{\det \widehat{\mathbf{B}}}$$
may be a useful (numerical) measure of how orthogonal a given basis $\mathbf{B}$ is.
This quantity is known as the orthogonality defect, and (by Hadamard's inequality) is lower-bounded by 1.
A natural interpretation of your question is then
How bad can the orthogonality defect of an LLL-reduced basis be?
To this end, there are some bounds known, but I do not recall how tight they are (I remember someone's thesis --- perhaps Phong Nyguen's or Damien Stehle's --- containing some discussion of this).
Still, the following upper-bound is known.
If $\mathbf{B}$ is a LLL-reduced basis of an $n$-dimensional lattice, then its orthogonality defect is at most $2^{n^2/2}$.
See for example Lemma 1.18, but this is a fairly well-known result (once you know the term "orthogonality defect" to search on).
