Direct Correlation (DC) Time Delay Estimation: Variance Keeps Decreasing for Increasing SNR?
I can reproduce a plateau value of about $0.0013$, compatible with the figure, as the error due to parabolic interpolation (2nd degree polynomial interpolation) around the peak of noise-free cross-correlation.

Without loss of generality, we assume $T=1$ and a uniform distribution of the delay $d$ between $-0.5$ and $0.5.$ Also we assume a perfectly white critically sampled signal and its delayed copy. Their cross-correlation is the critical samples of a time-shifted sinc function. We do a parabolic fit on three successive samples around the sample with the largest value. Calculating the peak of the parabola will estimate the delay as (calculations omitted):

$$\hat d = \frac{d}{2-4d^2}\tag{1}$$

Figure 1. Estimate $\hat d$ and true delay $d.$

The estimate is unbiased, as the expected value of its error is zero due to the symmetry seen in Fig. 1:

$$\operatorname{E}\left[\hat d - d\right] = 0$$

The variance of the error of the estimate will be:

$$\operatorname{Var}[\hat d - d] = \int_{-0.5}^{0.5}\left(\frac{d}{2-4d^2} -
 d - \operatorname{E}[\hat d - d]\right)^2dd \approx 0.0072$$

If we treat the left and right half-ranges of $d$ separately, the estimate is actually biased. The error can be reduced by subtracting from the estimate the expected value of the error conditional on the sign of $d$:

$$E\left[\hat d - d\,|\,-0.5 \le d &lt; 0\right] = 2\int_{-0.5}^0\left(\frac{d}{2 -
 4 d^2} - d\right)dd = -\frac{\ln(2)}{4} + \frac{1}{4}\approx 0.07671320485\\
E\left[\hat d - d\,|\,0 \le d &lt; 0.5\right] = \frac{\ln(2)}{4} - \frac{1}{4}$$

$$\hat{d_2} = \begin{cases}\hat d + \frac{\ln(2)}{4} - \frac{1}{4}&amp;\text{if }-0.5 \le d &lt; 0,\\
\hat d - \frac{\ln(2)}{4} + \frac{1}{4}&amp;\text{if }0 \le d &lt; 0.5.\end{cases}$$

Figure 2. Estimate $\hat{d_2}$ and true delay $d.$

$$\operatorname{E}\left[\hat {d_2} - d\right] = 0$$

$$\operatorname{Var}[\hat {d_2} - d] = \int_{-0.5}^{0.5}\left(\hat{d_2} -
 d - \operatorname{E}[\hat {d_2} - d]\right)^2dd \approx 0.0013$$

In practice you just add the constant $\pm 0.07671320485$ to the estimate $\hat d$ of the same sign, giving you $\hat{d_2}$.

Your signal is low-pass and not critically sampled so a parabolic estimate of the peak of cross-correlation will be more accurate even without any such correction.

For a sinc peak shape, Eq. 1 can be inverted and used to remove the bias in the estimate.