In principle, a computer can compute $\tau$ for any knot, using grid diagrams (or variants thereof); Baldwin and Gillam computed $\tau$ for knots up to 11 crossings a while ago. However, this is not practical. Unlike with Khovanov homology, there are no programs that can compute even $\widehat{HFK}$ for "large" knots (say, more than 20 crossings).

There are a couple of tricks you can use, that sometimes help:

• Plamenevskaya gave a lower bound on $\tau$ using the self-linking number of any representative of $K$; more precisely, if $T$ is a transverse representative of $K$, then $$ {\rm sl}\,T \le 2\tau(K) - 1.$$ Of course you can use it for both $K$ and its inverse, and see what you get. To generate some random Legendrian/transverse representatives of $K$ you can use Gridlink.

• Another lower bound is given in terms of the unknotting number (I believe this is due to Ozsváth and Szabó): $$ u(K) \le \tau(K),$$ and in fact there are more refined versions of this with signed unknotting numbers (for which I'm afraid I have no references, but I can try and dig them up).

• Of course, you can always try and simplify your knot using concordances, since $\tau$ is a concordance invariant (and then combine it with the previous tricks).

• Upper bounds depend a lot on the kind of information you have. Plamenevskaya's inequality also happens to give an upper bound (since it gives a lower bound on $\tau$ and $-\tau$); the Seifert and slice genus are obvious candidates, but the latter is usually quite hard to get your hands on.

In principle, a computer can compute $\tau$ for any knot, using grid diagrams (or variants thereof); Baldwin and Gillam computed $\tau$ for knots up to 11 crossings a while ago. However, this is not practical. Unlike with Khovanov homology, there are no programs that can compute even $\widehat{HFK}$ for "large" knots (say, more than 20 crossings). I have heard that Zoltán Szabó has a program that can compute $\widehat{HFK}$ for large knots (maybe around 50 crossings) in a matter of minutes; I don't think he has made it public yet, though, and I don't know if it computes $\tau$.

There are a couple of tricks you can use, that sometimes help:

• Plamenevskaya gave a lower bound on $\tau$ using the self-linking number of any representative of $K$; more precisely, if $T$ is a transverse representative of $K$, then $$ {\rm sl}\,T \le 2\tau(K) - 1.$$ The inequality is in fact an equality if $T$ is the closure of a quasi-positive braid (i.e. the closure of a braid that's a product of conjugates of positive generators). Of course you can use it for both $K$ and its inverse, and see what you get. To generate some random Legendrian/transverse representatives of $K$ you can use Gridlink.

• Another lower bound is given in terms of the unknotting number (I believe this is due to Ozsváth and Szabó): $$ u(K) \le \tau(K),$$ and in fact there are more refined versions of this with signed unknotting numbers (for which I'm afraid I have no references, but I can try and dig them up).

• Of course, you can always try and simplify your knot using concordances, since $\tau$ is a concordance invariant (and then combine it with the previous tricks).

• Upper bounds depend a lot on the kind of information you have. Plamenevskaya's inequality also happens to give an upper bound (since it gives a lower bound on $\tau$ and $-\tau$); the Seifert and slice genus are obvious candidates, but the latter is usually quite hard to get your hands on.

EDIT: added something about computing $\widehat{HFK}$ and the relationship between $\tau$ and the self-linking number.