One approach is to observe that $$\|u_0\|_{L^{\infty}(B_1^+)} \geq \frac{1}{16n}|f(0)|.$$ It suffices by linearity to prove this when $f(0) = 1$. Consider the barrier $$b(x) = \frac{1}{2n}\left(\left|x - \frac{1}{2}e_n\right|^2 - \frac{1}{8}\right).$$ Since $b \geq \frac{1}{16n}$ on $\partial B_1^+$, either $u_0 \geq b$ somewhere on $\partial B_1^+$ and we are done, or $u_0 \leq b$ on $\partial B_1^+$ in which case the comparison principle implies that $$u\left(\frac{e_n}{2}\right) \leq -\frac{1}{16n}$$ and we are again done.

With this estimate in hand, your suggestion to use reflection and interior estimates for the harmonic function $v_0 = u_0 - \frac{f(0)}{2}x_n^2$ works because $$\|v_0\|_{L^{\infty}(B_1^+)} \leq (8n+1)\|u_0\|_{L^{\infty}(B_1^+)}.$$

One approach is to observe that $$\|u_0\|_{L^{\infty}(B_1^+)} \geq \frac{1}{16n}|f(0)|.$$ It suffices by linearity to prove this when $f(0) = 1$. Consider the barrier $$b(x) = \frac{1}{2n}\left(\left|x - \frac{1}{2}e_n\right|^2 - \frac{1}{8}\right).$$ Since $b \geq \frac{1}{16n}$ on $\partial B_1^+$, either $u_0 \geq b$ somewhere on $\partial B_1^+$ and we are done, or $u_0 \leq b$ on $\partial B_1^+$ in which case the comparison principle implies that $$u\left(\frac{e_n}{2}\right) \leq -\frac{1}{16n}$$ and we are again done.

With this estimate in hand, your suggestion to use reflection and interior estimates for the harmonic function $v_0 = u_0 - \frac{f(0)}{2}x_n^2$ works because $$\|v_0\|_{L^{\infty}(B_1^+)} \leq (1+8n)\|u_0\|_{L^{\infty}(B_1^+)}.$$ More precisely, $$\|D^3u_0\|_{L^{\infty}(B_{1/2}^+)} = \|D^3v_0\|_{L^{\infty}(B_{1/2}^+)} \leq C(n)\|v_0\|_{L^{\infty}(B_1^+)} \leq C(n)(1+8n)\|u_0\|_{L^{\infty}(B_1^+)}.$$

One approach is to observe that $$\|u_0\|_{L^{\infty}(B_1^+)} \geq \frac{1}{16n}|f(0)|.$$ It suffices by linearity to prove this for $f(0) = 1$. Consider the barrier $$b(x) = \frac{1}{2n}\left(\left|x - \frac{1}{2}e_n\right|^2 - \frac{1}{8}\right).$$ Since $b \geq \frac{1}{16n}$ on $\partial B_1^+$, either $u_0 \geq b$ somewhere on $\partial B_1^+$ and we are done, or $u_0 \leq b$ on $\partial B_1^+$ in which case the comparison principle implies that $$u\left(\frac{e_n}{2}\right) \leq -\frac{1}{16n}$$ and we are again done.

With this estimate in hand, the approach of considering $v_0 = u_0 - \frac{f(0)}{2}x_n^2$ works because $$\|v_0\|_{L^{\infty}(B_1^+)} \leq (8n+1)\|u_0\|_{L^{\infty}(B_1^+)}.$$

One approach is to observe that $$\|u_0\|_{L^{\infty}(B_1^+)} \geq \frac{1}{16n}|f(0)|.$$ It suffices by linearity to prove this when $f(0) = 1$. Consider the barrier $$b(x) = \frac{1}{2n}\left(\left|x - \frac{1}{2}e_n\right|^2 - \frac{1}{8}\right).$$ Since $b \geq \frac{1}{16n}$ on $\partial B_1^+$, either $u_0 \geq b$ somewhere on $\partial B_1^+$ and we are done, or $u_0 \leq b$ on $\partial B_1^+$ in which case the comparison principle implies that $$u\left(\frac{e_n}{2}\right) \leq -\frac{1}{16n}$$ and we are again done.

With this estimate in hand, your suggestion to use reflection and interior estimates for the harmonic function $v_0 = u_0 - \frac{f(0)}{2}x_n^2$ works because $$\|v_0\|_{L^{\infty}(B_1^+)} \leq (8n+1)\|u_0\|_{L^{\infty}(B_1^+)}.$$