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The answer is no. Indeed, consider the case when $n=2$ and $$f(x,y)=\max(| x|,2 | y|)$$ for all $(x,y)\in\mathbb R^2$.

Then for $(x,y)$ near $(\pm1,0)$, we have $f(x,y)=|x|$ and hence $\partial f(\pm1,0)\subseteq B_1(0)$. On the other hand, $f(0,y)=2|y|$, so that $\partial f(0,0)\not\subseteq B_1(0)$.

The answer is no. Indeed, consider the case when $n=2$ and $$f(x,y)=\max(| x|,2 | y|)$$ for all $(x,y)\in\mathbb R^2$.

Then for $(x,y)$ near $(\pm1,0)$ we have $f(x,y)=|x|$ and hence $\partial f(\pm1,0)\subseteq B_1(0)$. On the other hand, $f(0,y)=2|y|$, so that $\partial f(0,0)\not\subseteq B_1(0)$.

The answer is no. Indeed, consider the case when $n=2$ and, over the rectangle $R:=[-1,1]\times[-h,h]$, the function $f$ is the pointwise maximum of the set of all affine functions $g$ such that $g(0,0)\le-1$, $g(0,h)\le-1+2h$, $g(0,-h)\le-1+2h$, $g(1,h)\le0$, $g(1,-h)\le0$, $g(-1,h)\le0$, and $g(-1,-h)\le0$, where $h>0$ is small enough, with $f$ appropriately extended outside the rectangle $R$.

Explicitly, for $h=1/10$, we have $$f(x,y)=\max\Big(| x| -1,\frac{4 | x| }{5}+2 | y| -1\Big)$$ for $(x,y)\in R$, and then extend $f$ from $R$ to $\mathbb R^2$ by the formula $$f(x,y)=\sup\{f(u,v)+p\cdot(x-u,y-v)\colon \\ (u,v)\in(-1,1)\times(-h,h), p\in\partial f(u,v)\} \\ =\max\Big(| x| -1,\frac{4 | x| }{5}+2 | y| -1\Big)$$ actually for all $(x,y)\in\mathbb R^2$, where $\cdot$ denotes the dot product.

Then for $(x,y)$ near $(\pm1,0)$, we have $f(x,y)=|x| -1$ and hence $\partial f(\pm1,0)\subseteq B_1(0)$. On the other hand, $f(0,y)=-1+2|y|$ for small enough $|y|$, so that $\partial f(0,0)\not\subseteq B_1(0)$.

The answer is no. Indeed, consider the case when $n=2$ and $$f(x,y)=\max(| x|,2 | y|)$$ for all $(x,y)\in\mathbb R^2$.

Then for $(x,y)$ near $(\pm1,0)$, we have $f(x,y)=|x|$ and hence $\partial f(\pm1,0)\subseteq B_1(0)$. On the other hand, $f(0,y)=2|y|$, so that $\partial f(0,0)\not\subseteq B_1(0)$.

The answer is no. Indeed, consider the case when $n=2$ and, over the rectangle $R:=[-1,1]\times[-h,h]$, the function $f$ is the pointwise maximum of the set of all affine functions $g$ such that $g(0,0)\le-1$, $g(0,h)\le-1+2h$, $g(0,-h)\le-1+2h$, $g(1,h)\le0$, $g(1,-h)\le0$, $g(-1,h)\le0$, and $g(-1,-h)\le0$, where $h>0$ is small enough, with $f$ appropriately extended outside the rectangle $R$.

Explicitly, for $h=1/10$, $$f(x,y)=\max\left(| x| -1,\frac{4 | x| }{5}+2 | y| -1\right)$$ for $(x,y)\in R$, with $$f(x,y)=\sup\{f(u,v)+p\cdot(x-u,y-v)\colon \\ (u,v)\in(-1,1)\times(-h,h), p\in\partial f(u,v)\}$$ for all $(x,y)\in\mathbb R^2$, where $\cdot$ denotes the dot product.

Then for $(x,y)$ near $(\pm1,0)$, we have $f(x,y)=|x| -1$ and hence $\partial f(\pm1,0)\in B_1(0)$. On the other hand, $f(0,y)=-1+2|y|$ for small enough $|y|$, so that $\partial f(0,0)\notin B_1(0)$.

The answer is no. Indeed, consider the case when $n=2$ and, over the rectangle $R:=[-1,1]\times[-h,h]$, the function $f$ is the pointwise maximum of the set of all affine functions $g$ such that $g(0,0)\le-1$, $g(0,h)\le-1+2h$, $g(0,-h)\le-1+2h$, $g(1,h)\le0$, $g(1,-h)\le0$, $g(-1,h)\le0$, and $g(-1,-h)\le0$, where $h>0$ is small enough, with $f$ appropriately extended outside the rectangle $R$.

Explicitly, for $h=1/10$, we have $$f(x,y)=\max\Big(| x| -1,\frac{4 | x| }{5}+2 | y| -1\Big)$$ for $(x,y)\in R$, and then extend $f$ from $R$ to $\mathbb R^2$ by the formula $$f(x,y)=\sup\{f(u,v)+p\cdot(x-u,y-v)\colon \\ (u,v)\in(-1,1)\times(-h,h), p\in\partial f(u,v)\} \\ =\max\Big(| x| -1,\frac{4 | x| }{5}+2 | y| -1\Big)$$ actually for all $(x,y)\in\mathbb R^2$, where $\cdot$ denotes the dot product.

Then for $(x,y)$ near $(\pm1,0)$, we have $f(x,y)=|x| -1$ and hence $\partial f(\pm1,0)\subseteq B_1(0)$. On the other hand, $f(0,y)=-1+2|y|$ for small enough $|y|$, so that $\partial f(0,0)\not\subseteq B_1(0)$.