[x1 , x2 , x3 , x4 , x5 , x6] = vpasolve( eqn,symvar(eqn) )
if you breakpoint at that line, and run to there, then you can do
sol1_4 = solve(eqn(1:4), [T_p1, T_p2, T_p3, T_p4])
neqn = simplify(subs(eqn(5:end), sol1_4))
It is then possible to solve neqn(1) for T_p5 -- or at least Maple can do it. The result is T_p5 expressed in terms of the roots of a polynomial of degree 56 with non-zero coefficients at degree 56, 50, 31, 25, and 0.
Now you could hope for a moment to factor into a polynomial of degree 31 times a polynomial of degree 25, but that would get you entries with degree 56, 31, 25, and 0... but no entries of degree 50. And if you try to include (polynomial of degree 25)^2 then you will start to get additional powers; conjugate roots can get you degree 50 but not the 25 needed to add to 31 to get 56. Anyhow, you probably cannot factor.
Once you have the root() form of the polynomial of degree 56, you can substitute it into neqn(2) to get the equation that needs to hold for T_p6 . You are not going to be able to solve that symbolically. If you try to solve it numerically... you have problems. If you plot the imaginary part, it looks like there might be only one place the imaginary part is 0, near 20.749... but at that location, the real part is over 100000.
It appears that all of the roots for T_p6 are imaginary.