It seems to depend on the strength assumed of the convergence of the $\phi_k$. This is a partial answer, where in some parts it is assumed that $p > n - 1$. Under this hypothesis, the convergence does not hold if only $\lvert \phi_k \rvert_{L^\infty(\partial B_1)} \to 0$. However if $\lvert \phi_k \rvert_{C^{0,\beta}(\partial B_1)} \to 0$ for some $\beta \in (1 - \frac{1}{p},1)$ then also $\lvert v_k \rvert_{W^{1,p}(B_1)} \to 0$, regardless of the relative size of $p$ and $n$.
The trace operator is instrumental in seeing this: recall that there is a constant $C > 0$ so that $\lvert \mathrm{tr} \, u \rvert_{W^{1-1/p,p}(\partial B_1)} \leq C \lvert u \rvert_{W^{1,p}(B_1)}$ for all $u \in W^{1,p}(B_1)$. Here and throughout $W^{1-1/p,p}(\partial B_1)$ is a fractional Sobolev-Slobodeckij space. Recall also that there is an extension operator $W^{1,1-1/p}(\partial B_1) \to W^{1,p}(B_1)$ that is a right inverse to the trace.
In particular, if there were the strong convergence $\lvert v_k \rvert_{W^{1,p}(B_1)} \to 0$ then also $\lvert \phi_k \rvert_{W^{1-p,p}(\partial B_1)} \to 0$. This is strictly stronger than $L^\infty$-convergence because $W^{1-1/p,p}(\partial B_1)$ embeds into the Holder space $C^{0,\alpha}(\partial B_1)$, where $\alpha \in (0,1 - \frac{1}{p} -\frac{n-1}{p}] = (0,1-\frac{n}{p}]$. Therefore, if one chose a sequence of traces so that $\lvert \phi_k \rvert_{C^{0,\alpha}(\partial B_1)} \not\to 0$ for some exponent $\alpha$ in this range then the convergence $\lvert v_k \rvert_{W^{1,p}(B_1)} \to 0$ would be impossible. (This is where the hypothesis $p > n - 1$ is used: when $p < n - 1$ then instead $W^{1-p,p}(\partial B_1) \subset L^s(\partial B_1)$ for all $s \in (0,\frac{(n-1)p}{n-1-p}]$.)
However, if $\beta > 1 - \frac{1}{p}$ and one assumes $\lvert \phi_k \rvert_{C^{0,\beta}(\partial B_1)} \to 0$ then the convergence $\lvert v_k \rvert_{W^{1,p}(B_1)} \to 0$ is guaranteed. This is because the 'reverse' inclusion $C^{0,\beta}(\partial B_1) \subset W^{1-1/p,p}(\partial B_1)$ holds. Therefore $\phi_k \in W^{1-1/p,p}(\partial B_1)$; let $u_k \in W^{1,p}(B_1)$ be its image under the extension operator. By the above there is $C > 0$ so that $\lvert u_k \rvert_{W^{1,p}(B_1)} \leq C \lvert \phi_k \rvert_{C^{0,\beta}(\partial B_1)}$ for all $k$. By minimality, the same holds with $v_k$ replacing $u_k$, and therefore $\lvert v_k \rvert_{W^{1,p}(B_1)} \to 0$ also. (Note that the assumption that $p > n-1$ is not needed here.)