A harmonic function

Yes, it can be found explicitly, though not in elementary functions but in terms of a combination of elementary and hypergeometric functions. The problem is (almost) equivalent to finding a conformal map from your straight strip onto the broken strip. For this it is sufficient to find a conformal map from the upper half-plane onto the UPPER half of the broken strip (and then apply the Schwarz Symmetry principle). This map of the upper half-plane on the upper half of the broken strip is given by the Christoffel-Schwarz formula $$C\int_0^z\zeta^{\beta-1}(\zeta-T)^{-\beta} d\zeta,$$ where $\alpha=\pi\beta$. This integral can be explicitly expressed in terms of hypergeometric function. I hope these hints are enough to make an explicit computation.

In a comment you ask about dependence on $T$. Dependence on $T$ is trivial: If $F$ is the conformal mapping corresponding to $T=1$, normalized such that $F(0)=0,F(1)=1,F(\infty_j)=\infty_j, j=1,2$ then $TF(z/T)$ is the map corresponding to $T$.

Remark. According to a theorem of Chebyshev, this integral is not an elementary function when $\beta$ is irrational. Therefore I suppose that the solution of your problem is also not an elementary function.

Here are some more detail. Let $T=1$. Let $F(z)$ be the integral I wrote above. Then $C$ is determined from the boundary correspondence: $C=1/F(1)$. Now $$G(z)=(-\cos(\pi z)+1)/2$$ maps the rectangular half-strip onto the upper half-plane, $(0,1,\infty)\mapsto (0,1,\infty)$. Therefore $$H(z)=F(G(z))/F(1)$$ maps the upper half of the rectilinear strip on the upper half of the slanted strip (both with $T=1$). So the solution of the Dirichlet problem (with $T=1$) is $$\Re(H^{-1}(z)).$$ Not very explicit, of course, since one has to invert the integral $F$. But OK for numerical computation.

Remark 2. The integral $$B_x(\beta,1-\beta):=\int_0^x\zeta^{\beta-1}(1-\zeta)^{-\beta}d\zeta$$ is called the incomplete Beta-function, and there is even the standard notation $$I_x(\beta,1-\beta):=\frac{B_x(\beta,-\beta)}{B(\beta,1-\beta)}=F(x)/F(1).$$ To obtain simple estimates, expand it to the power series: $$B_x(\beta,1-\beta)=\beta^{-1}x^\beta+\sum_{n=1}^\infty\frac{\beta(\beta+1)\ldots(\beta+n-1)}{n!(n+\beta)}x^{n+\beta}.$$ This is sufficient to determine your function on $(0,c)$ with any desired accuracy, unless $c$ is close to $1$. Near $1$ use a similar expansion into powers of $x-1$.

For example, near $0$, $G(x)\sim x^2/4,$ so $$H(x)\sim\frac{1}{\beta 4^{\beta}B(\beta,1-\beta)}x^{2\beta},$$ $H^{-1}(x)\sim Cx^{1/(2\beta)}.$ Also notice that the series for $B_x$ has positive coefficients which simplifies the estimates.

To obtain a lower estimate, you need the upper estimate to $B_x(\beta,1-\beta)$. We can write it as $$B(x):=B_x(\beta,1-\beta)=\beta^{-1}x^\beta+f(x),$$ where $f$ is increasing (because it is a power series with positive coefficients), and we also know that $B(1)=B(\beta,1-\beta),$ therefore $f(x)<f(1)=B(\beta,1-\beta)-1/\beta$. This will give an explicit constant $C$ in the lower estimate $Cx^{(1/2\beta)}$.