It looks like the test you are looking at is a psychometric test, which uses a different T-score than what you would get from a T-test. T-scores in psychometric testing is just a scaled z-score. The formula is T = (Z x 10) + 50, with Z being the z-score and T being the T-score.

According to Springer article on T-Score

T scores have a mean of 50 and a standard deviation of 10. Standard z scores can be converted to T scores using the formula below.

T=10∗z+50

So using MathNet.Numerics package you can simply do

    static double PercintileFromTScore(double tScore)
    {
        var normalDistribution = new MathNet.Numerics.Distributions.Normal(50, 10); // shoulbe be shorten "using" in the real life
        return normalDistribution.CumulativeDistribution(tScore);
    }

This seems to produce results identical to the table you provided.

It is possible to calculate the percentile given a t-statistic and the degrees of freedom. This is the CDF at the given t-value.

The formula for this is as follows, where $\nu$ is the degrees of freedom, $t^{2} < \nu $, and $_{2}F_{1}$ is a hypergeometric function.

$\int^{t}_{-\inf} f(u) du = \frac{1}{2} + t [ \frac{\Gamma(\frac{\nu + 1}{2})}{\sqrt{\pi\nu}\Gamma(\frac{\nu}{2})}] [ _{2}F_{1}(\frac{1}{2}, \frac{\nu + 1}{2}; \frac{3}{2}; -\frac{t^2}{\nu})]$

This can be easily done in R using pt().