Any computable transcendental number that is computable in P time but not $O(n)$
Here is the construction of such a number.  You can argue whether this means such a number is "known".

Take any function $f$ from $\mathbb{N}$ to $\{ 1, 2, \ldots, 8 \}$ where the $n$'th digit is not computable in $O(n)$ time.  Such a function exists, for example, by the usual diagonalization technique.   Interpret $f(n)$ as the $n$'th decimal digit of some real number $\alpha$.   Now, for each $n$ of the form $2^{2^k}$, $k \geq 1$, change the digits of $\alpha$ in positions $n, n+1, \ldots, 3n$ to $0$'s.   The resulting number $\beta$ evidently retains the property that the $n$'th digit is not computable in $O(n)$ time, but has infinitely many very good approximations by rationals, say to order $O(q^{-3})$, of the form $p/q$.  Then by Roth's theorem $\beta$ cannot be algebraic.  (It is not rational because it has arbitrarily long blocks of $0$'s set off by nonzeros on both sides.)