Difference between PSD estimate and variance of DFT
So, as usual, I start by pointing out what the PSD actually is:

The power spectral density is the Fourier Transform of the autocorrelation function of a wide-sense stationary random process.


  Always assuming  $\hat{X}=0$, is it generally correct  to interpret the PSD as an biased estimate of the DFT variance? If not, can someone  explain  the difference, please?


The Wiener-Chintschin¹ theorem points out that this happens to coincide with the expectation value of the absolute of the Fourier transform of a signal.  In other words, to answer your first question:

The PSD is a stochastic (hence, "invisible") property of a random process. The DFT can be used to estimate that; not the other way around!

 

To answer your second question, a bit of juggling the term of variance is required:


  Would Bartlett's method become an unbiased estimator for the DFT variance if (3) was scaled by $\frac{1}{L-1}$ instead of $\frac{1}{L}$?


Variance is formally defined as the expectation of the square of the deviation of any observation to the mean:

$$Var(X) = \mathbf E\left[\left(X-\mu\right)^2\right]$$

Now, applying Parseval's theorem, the power in a time and frequency domain should be identical. Considering a white random process, where the covariance of two samples is 0, it's easy to see that variance is kind-of-linear:

$$Var\left(\sum\limits_{i=1}^L X_i\right) = \sum\limits_{i=1}^N Var(X_i)$$

Considering the special case of L=1, it becomes clear that $(4)$ isn't applicable here – you kind of get the standard deviation of the process, but not the variance.


¹ there's more latin-lettered spellings to Chintschin than there's letters in in his cyrillic name...