Yes, there are other spaces in which Borel's Theorem holds (property (BT)).

To see this, note that every cartesian product of locally convex spaces with (BT) has (BT). In particular, any uncountable power V:=R^I of the field of real numbers has (BT). Yet, this space cannot be given a finer Frechet space topology S.

In fact, for any such topology S, we could choose a decreasing sequence of
balanced 0-neighbourhoods U_n in (V,S) which form a basis of 0-neighbourhoods. Let C_n be the closure of U_n in R^I with respect to the product topology.

Now R^I (with the product topology) is a Baire space (see Oxtoby, J.C., Cartesian products of Baire spaces, Fundam. Math. 49 (1961), 157-166).

As in the usual proof of the open mapping theorem, we see that each C_n
is a 0-neighbourhood in the cartesian product R^I. Since each 0-neighbourhood in the cartesian product R^I contains some C_n, we see that the cartesian product R^I would be first countable, which is absurd.

This contradiction shows that S cannot exist.