[I've corrected a stupid mistake below and added an upper bound... Please check the numerical values!]

Well, I doubt that an explicit expression exists. However, it should be possible to get good bounds. The lower bound is easy: observe that
$$
E(T) = \sum_{k\geq 1} P(T> k-1) = 1+\sum_{k\geq 1} c_k 4^{-k},
$$
where $c_k$ is the number of self-avoiding paths of length $k$ (and, of course, $4^k$ is the number of all paths of length $k$), so that we get a lower bound by truncating this series. Using the (known) values for $c_k$, $k=1,\ldots,71$, we get
$$
E(T) > 4.58607909
$$
Now, you can get an upper bound by bounding the neglected part of the series using $c_k\leq 4 \cdot 3^{k-1}$. This gives you a very narrow interval containing the right value: if I have made no mistake ;) , we get
$$
4.58607909 < E(T) < 4.58607911
$$

5more comments