A rough estimate is obtained as follows. The $L^\infty$ norm$L^{+\infty}$−norm of $S(t)y$ is majorized by $\sum_{n=1}^\infty e^{-n^2\pi^2 t}|\langle y,\sin(n\pi x)\rangle|$ which in turn is majorized by $\|y\|_{L^1}\sum_{n=1}^\infty e^{-n^2\pi^2 t}$ assuming that $\langle y,\sin(n\pi x)\rangle$ is intended to mean $\int_0^1y(s)\sin(n\pi s){\kern.7mm\rm d\kern.7mm}s$. So it remains to estimate $\sum_{n=1}^\infty e^{-n^2\pi^2 t}$. This is majorized by $\int_0^{+\infty}e^{-s^2\pi^2 t}{\kern.7mm\rm d\kern.7mm}s$. Recalling that $\int_0^{+\infty}e^{-u^2}{\kern.7mm\rm d\kern.7mm}u=\frac12\sqrt\pi$, and making the substitution $u=s\pi\sqrt t$, we obtain $\pi\sqrt t\int_0^{+\infty}e^{-s^2\pi^2 t}{\kern.7mm\rm d\kern.7mm}s=\frac12\sqrt\pi$ whence further $\int_0^{+\infty}e^{-s^2\pi^2 t}{\kern.7mm\rm d\kern.7mm}s=t^{-\frac12}\frac1{2\sqrt\pi}$. So $C=\frac1{2\sqrt\pi}$ is one possibility.
By the way, if one wants to have $S(0)y=y$ for at least $y=\langle\kern.6mm\sin(n\pi s):0<s<1\kern.6mm\rangle$ with $n\in\mathbb Z^+$, one should instead have $$ (S(t)y)(s)=\sum_{n=1}^{+\infty}e^{-n^2\pi^2 t}\left(2\int_0^1y(s)\sin(n\pi s){\kern.7mm\rm d\kern.7mm}s\right)\kern.7mm\sin(n\pi s)\text{ .} $$ Then my argument above gives $C=\pi^{-\frac12}\kern.6mm$.