I'm trying to figure out why higher degrees of freedom $(n-1-k)$ in a linear regression is "better". I can't see how higher df would automatically result in lower $MSE$, since every new df (data point) increases the sum of squared errors in the numerator, just as it increases $n-1-k$ in the denominator. It would make sense if there was some variance measure like $MSE / n$, where higher df would unambiguously reduce variance, just like for a univariate distribution where increasing df reduces the variance of the sample mean.
In linear regression, if you force the slope coefficient to be $0$, it reduces to a univariate model, $Y = \beta_0 + error$. $\beta_0$ is the sample mean of the $y$ values. The sample variance = sum of squared errors / $(n-1)$. And the variance of the sample mean = sample variance / $n$.
Now say you allow the slope parameter to be nonzero, i.e. $Y = \beta_0 + \beta_1*X + error$. The mean squared error of the estimate ($MSE$) = sum of squared errors / $(n-1-k)$. From this, I assume $MSE$ is analogous to sample variance in the above no‐slope model, since they both refer to the variance of the error term.
So is there some concept for linear regression like the variance of the sample mean error = $MSE$ / $n$? Something that would be analogous to the variance of the sample mean = sample variance / $n$ (in a univariate setting)? Or is it meaningless because the mean residuals of the regression in any sample will by definition be equal to $0$, i.e. the expected value of the error?