Norm of the linear functional $f\mapsto \int_0^1 f(t)/t dt$

You can split the integral in the following way to get a better bound. Let $a\in(0,1)$, then

$$ \left\lvert\int_{0}^{1} \frac{f(t)}{t}dt\right\rvert\leq\left\lvert\int_{0}^{a}\frac{f(t)}{t}dt\right\rvert+\left\lvert\int_{a}^{1}\frac{f(t)}{t}dt\right\rvert. $$ Now the second term clearly satisfies$$ \left\lvert\int_{a}^{1}\frac{f(t)}{t}dt\right\rvert\leq \lVert f\rVert_\infty \int_{a}^{1}\frac 1t dt=\lVert f\rVert_\infty (-\ln(a)). $$ The first term on the other hand satisfies, by your own argument with integration by parts $$ \left\lvert\int_{0}^{a} \frac{f(t)}{t}dt\right\rvert\leq \left\lvert\int_{0}^{a}\left(\ln(a)-\ln(t)\right) f'(t)dt\right\rvert\leq a \lVert f'\rVert_\infty. $$ Hence you have $\lvert \varphi(f)\rvert\leq \max(a,-\ln(a))\lVert f \rVert_{C^1}$ for any $a\in(0,1)$. Optimizing in $a$, you find the minimum value of $\min_{a\in(0,1)}\max(a,-\ln(a))\approx 0.567$.