Yes, your bound is correct. Let $X = C[0,1]$ with the $\sup$-norm and let $\{\alpha_k\}_{k = 1}^n \subset [0,1]$ as in the question. Let $f \in X$ be arbitrary with $\Vert f \Vert = 1$, then
$$ \vert Tf \vert = \vert \sum_{k = 1}^n \alpha_k f(x_k) \vert \leq \sum_{k = 1}^n \vert \alpha_k \Vert f \Vert_{\infty} \vert = \sum_{k = 1}^n \vert \alpha_k \vert $$
and hence $$\Vert T \Vert = \sup_{f \in X, \Vert f \Vert = 1} \vert Tf \vert \leq \sum_{k = 1}^n \vert \alpha_k \vert ~~.$$
In fact, this is the norm of the functional, since $\sum_{k = 1}^n \vert \alpha_k \vert$ is attained by the following function $f$. Define
$$ f(x) = \begin{cases} 1 & x = x_k, \alpha_k \geq 0 \\
-1 & x = x_k, \alpha_k < 0 \\
0 & x \in \{0,1\} \setminus \{x_k\}_{k =1}^n
\end{cases}$$
and linear interpolation otherwise. Then this function is continuous on $[0,1]$, it has $\Vert f \Vert = 1$, and we have
$$ \vert Tf \vert = \vert \sum_{k = 1}^n \alpha_k f(x_k) \vert = \sum_{k = 1}^n \vert \alpha_k \vert$$
where in the last step we use that all summands are positive. Hence $\Vert T \Vert \geq \sum_{k = 1}^n \vert \alpha_k \vert$, and thus we have equality.
Too long to be a comment:
This is just an elementary upper bound.
We have
\begin{align*}
\Vert Tf \Vert_{2}^2 &= \int_0^1 \vert Tf(x)\vert^2 dx
=\int_0^1 \left\vert \int_0^1 \vert \sin(x-y)\vert^{-\alpha} f(y) dy \right\vert^2 dx \\
&\leq \int_0^1 \left( \int_0^1 \vert \sin(x-y)\vert^{-\alpha} \vert f(y)\vert dy \right)^2 dx \\
&= \int_0^1 \left(\int_0^1 \vert\sin(x-y_1)\vert^{-\alpha} \vert f(y_1) \vert dy_1\right)\left( \int_0^1 \vert\sin(x-y_2)\vert^{-\alpha} \vert f(y_2) \vert dy_2 \right) dx \\
&= \int_0^1 \int_0^1 \int_0^1 \vert\sin(x-y_1)\vert^{-\alpha} \vert f(y_1) \vert \cdot \vert\sin(x-y_2)\vert^{-\alpha} \vert f(y_2) \vert dy_1 dy_2 dx.
\end{align*}
Using Cauchy-Schwarz and then Tonelli's theorem we get
\begin{align*}
\Vert Tf \Vert_{2}^2 &\leq \left( \int_0^1 \int_0^1 \int_0^1 \vert\sin(x-y_1)\vert^{-\alpha} \vert\sin(x-y_2)\vert^{-\alpha} \vert f(y_1) \vert^2 dy_1 dy_2 dx\right)^{1/2} \\
&\qquad \times \left( \int_0^1 \int_0^1 \int_0^1 \vert\sin(x-y_1)\vert^{-\alpha} \cdot \vert\sin(x-y_2)\vert^{-\alpha} \vert f(y_2) \vert^2 dy_1 dy_2 dx\right)^{1/2} \\
&=\left( \int_0^1 \int_0^1 \int_0^1 \vert\sin(x-y_1)\vert^{-\alpha} \vert\sin(x-y_2)\vert^{-\alpha} \vert f(y_1) \vert^2 dy_2 dx dy_1\right)^{1/2} \\
&\qquad \times \left( \int_0^1 \int_0^1 \int_0^1 \vert\sin(x-y_1)\vert^{-\alpha} \cdot \vert\sin(x-y_2)\vert^{-\alpha} \vert f(y_2) \vert^2 dy_1 dx dy_2\right)^{1/2} \\
&=\int_0^1 \int_0^1 \int_0^1 \vert\sin(x-s)\vert^{-\alpha} \cdot \vert\sin(x-t)\vert^{-\alpha} \vert f(t) \vert^2 ds dx dt.
\end{align*}
After the change of variables $w=s-x$ we get
\begin{align*}
\Vert Tf \Vert_{2}^2 &\leq \int_0^1 \int_0^1 \int_{-x}^{1-x} \vert \sin(w)\vert^{-\alpha} \cdot \vert \sin(x-t) \vert^{-\alpha} \cdot \vert f(t)\vert^2 dw dx dt \\
&\leq \left( \sup_{v\in [0,1]} \int_{-v}^{1-v} \vert \sin(w)\vert^{-\alpha} dw \right) \int_0^1 \int_0^1 \vert \sin(x-t) \vert^{-\alpha} \cdot \vert f(t)\vert^2 dx dt.
\end{align*}
Repeating this last step for the integration in $x$ we get
\begin{align*}
\Vert Tf \Vert_2^2 \leq \left( \sup_{v\in [0,1]} \int_{-v}^{1-v} \vert \sin(w)\vert^{-\alpha} dw \right)^2 \Vert f \Vert_2^2.
\end{align*}
Hence, we get
\begin{align*}
\Vert T f\Vert_2 \leq \left( \sup_{v\in [0,1]} \int_{-v}^{1-v} \vert \sin(w)\vert^{-\alpha} dw \right) \Vert f \Vert_2
\end{align*}
and therefore
\begin{align*}
\Vert T \Vert \leq \sup_{v\in [0,1]} \int_{-v}^{1-v} \vert \sin(w)\vert^{-\alpha} dw.
\end{align*}
As the map
\begin{align*}
G:[0,1] \rightarrow \mathbb{R}, v \mapsto \int_{-v}^{1-v} \vert \sin(w)\vert^{-\alpha} dw
\end{align*}
is continuous and $[0,1]$ is compact, the maximum is attained at some point $v_0\in [0,1]$. In fact we have that the supremum is attained for $v_0=1/2$.
Best Answer
Approximate the function $$ f(x)=\begin{cases}1,&x\leq 1/2\\ -1,&x>1/2 \end{cases} $$ with continuous functions.
For this, we take the sequence of norm $1$ continuous functions: $$ f_n(x)=\begin{cases} 1,&x\leq 1/2-1/n\\ -n(x-1/2),&1/2-1/n<x\leq 1/2+1/n\\ -1,&1/2+1/n<x\leq 1 \end{cases} $$ I leave to you to verify $F(f_n)\uparrow 3/2$