Let $p_0, p_1, \ldots$ be the orthonormal polynomials for the measure $\mu$ on the interval $[a,b]$. That is, $\displaystyle \int_a^b p_i(x)p_j(x) d\mu(x) = \delta_{i,j}$ with $\delta_{i,j}$ the Kronecker delta. Define the Christoffel-Darboux kernel
$$K_n(x,y) = \sum_{j=0}^n p_j(x)p_j(y)$$
And define the operator
$$K_n(f; x) = \int_a^b f(y)K_n(x,y) d\mu(y)$$
I would like to find the two operator norms $||K_n||_{\infty}$ and $||K_n||_2$.
In both cases one can take $f(y)$ outside of the integral and bound it by $||f||_{\infty} \le \mu([a,b])^{1/2}||f||_2$ so we are left with bounding the norm of $\int_a^b K_n(x,y) d\mu(y)$.
To find the latter norm, we might want to use the Christoffel-Darboux formula which states that, for some constant $a_n$,
$$\sum_{j=0}^{n} p_j(x)p_j(y) = a_{n} \frac{p_{n+1}(x)p_{n}(y) - p_{n+1}(y)p_{n}(x)}{x - y}$$
And this holds for $x \neq y$. For $x = y$ we have
$$\sum_{j=0}^{n} p_j(x)p_j(y) = a_n \left(p'_{n+1}(x)p_n(x) - p_{n+1}(x)p'_n(x) \right)$$
However, I do not know how to proceed further.
I suspect you meant $\|f\|_2\le\mu([a,b])^{1/2}\|f\|_\infty$. But in any case, I'm not sure this is needed.
The $L^2$ case:
Observe that $\{p_n\}_0^\infty$ is an orthonormal basis of $L^2([a,b],\mu)$. Let $(,)$ be the inner product. By Bessel's inequality, we have $\sum_0^\infty|(f,p_i)|^2\le\|f\|^2_2$, so $$ \|K_n(f)\|_2^2=\sum_{i=0}^n|(f,p_i)|^2\le\sum_{i=0}^\infty|(f,p_i)|^2\le\|f\|^2_2, $$ i.e. $\|K_n\|_{L^2\to L^2}\le 1$. But $K_n(p_0)=p_0$, so $\|K_n\|_{L^2\to L^2}=1$.
The $L^\infty$ case:
It is easy to see that $$ \|K_n\|_{L^\infty\to L^\infty}\le \sup_{x\in[a,b]}\int_a^b|K_n(x,y)|d\mu(y). $$ In fact, this is an equality. Fix $x\in[a,b]$ and choose $f(y)=\text{sgn}\, K_n(x,y)$. Then $K_n(f;x)=\int_a^b|K_n(x,y)|d\mu(y)$, so by continuity, $\|K_n(f)\|_\infty\ge \int_a^b|K_n(x,y)|d\mu(y)$. Taking the supremum over $[a,b]$ gives the result.