This is getting much closer to the answer I was looking for. I will note that the vectors that represent the two balanced factors in the design matrix are not orthogonal. (Their inner product will be 20--the number of cases where both take value 1.) But the fact that they are uncorrelated means some transformation of those vectors will be orthogonal. So I think this is pretty close.

@user277126, rather than just a drive-by comment calling my question "baseless", it would be more helpful to post an answer showing the design matrix of the cell means model as an answer so I can see where I may have gone wrong. Also, for the record, the sum of squares for many, many ANOVA calculations I've done over the years show that the claim is not baseless. For example, Type I ANOVA only yields the same answers independent of entering the factors into the model (neatly partitioning the SS) when the sample sizes are equal across groups.

I'm not sure I completely buy either explanation. In the first half, I honestly don't see the connection you're trying to make with chi-square. There are no row sums, column sums, or table sums in ANOVA (unless I'm misunderstanding your point). And the second explanation isn't clear either. What does the regression line here have to do with ANOVA? Either way, I'm still looking for orthogonality of some vectors somewhere.

I've written down the design matrix and (1) the columns do not appear to be orthogonal, and (2) even if they were, I fail to see how equal sample sizes would contribute to that.

To my knowledge, you cannot trick brms into giving you frequentist inference. The number of chains and all that is irrelevant because MCMC algorithms are not inherently Bayesian or frequentist; they just sample from a probability distribution. The "guts" of a brm call sample from a posterior distribution and there's no way around that. In some cases, flat uniform priors give answers similar to frequentist answers, but I don't know that it's the case for multilevel models. And you're still not getting a frequentist answer...just a Bayesian one that kind of agrees with a frequentist one.

I am also confused by the claim about grades. Could you provide more context? Was this a quote from a larger passage? That might help clarify the meaning.

I'll just type it up now.

You can get a little more bang for your buck by running your chains in parallel.

I'll also add that the data set is somewhat large-ish (over 7000 observations).