Estimate Delay of a Known Signal Delayed by Sub Sample Resolution
Since the Matched Filter is used the Cross Correlation becomes "Auto Correlation" which is assured to be "Symmetric" relative to its maximum.

Hence a good approximation would be to approximate the area around the sampled peak by a Symmetric Function as a parametric model of the function.

The most trivial and easiest model for such function would be the Parabolic Function which is illustrated in the following.

Here's some scilab code that does the cross-correlation and then finds the inter-sample peak by doing a simple (ish) parabolic interpolation.

The plot shows the absolute relative error between the truth and the estimate.



// 25684
vals= [];
taus = [-0.5:.005:0.5] + 10^-9; // Add a small amount to not have a zero.

for tau = taus

    N = 1000;
    t = 0:N-1;
    omega=0.1902834909;
    phi = 2*%pi*0.3823483924;

    x = sin(omega*t + phi);
    x_tau = sin(omega*(t-tau) + phi);

    xc = xcorr(x_tau,x);

    [mx,ix] = max(xc);

    x_vals = [-1 0 +1];
    y_vals = xc(ix + x_vals);

    //https://stackoverflow.com/a/717791/12570
    den = (x_vals(1) - x_vals(2))*(x_vals(1) - x_vals(3))*(x_vals(2) - x_vals(3));
    A = x_vals(3)*(y_vals(2) - y_vals(1)) + x_vals(2)*(y_vals(1) - y_vals(3)) + x_vals(1)*(y_vals(3) - y_vals(2))
    A = A/den;

    B = x_vals(3)^2*(y_vals(1) - y_vals(2)) + x_vals(2)^2*(y_vals(3) - y_vals(1)) + x_vals(1)^2*(y_vals(2) - y_vals(3));
    B = B/den;

    C = x_vals(2) * x_vals(3) * (x_vals(2) - x_vals(3))* y_vals(1) + x_vals(3) * x_vals(1) * (x_vals(3) - x_vals(1))* y_vals(2) + x_vals(1) * x_vals(2) * (x_vals(1) - x_vals(2))* y_vals(3);
    C = C/den;

    x_v = -B / (2*A);
    y_v = C - B*B / (4*A);

    vals= [vals; tau x_v y_v];

end 

clf
plot(taus,abs((vals(:,2)' - taus)./taus));
xlabel('True value of delay');
title('Error in using parabola peak as subsample delay estimate');