The answer is the $p$-regularisation of $(1^n)$: this is $(1^n)$ if $n < p$ and otherwise the partition
$$ (r^a, (r-1)^{p-1-a}) $$
with $p-1$ parts, where $r$ and $a \in \{0,1,\ldots, p-1\}$ are defined uniquely by
$$(r-1)(p-1) + a = n, $$
taking $a = 0$ if $p-1$ divides $n$.
For more on this see G. D. James, On the decomposition matrices of the symmetric groups II, J. Algebra 43 (1976), 45--54. The Mullineux Conjecture (now proved) determines the partition labelling a general $D^\lambda \otimes \mathrm{sgn}$ for $\lambda$ a $p$-regular partition. The shortest proof I know is in Example 24.5(iii) in James' Springer lecture notes on the symmetric groups.
Edit. I'd like to record a slightly unusual proof here. By some general theory, the permutation module $M$ for a finite group $G$ acting on the cosets of $H \le G$ defined over a field $F$ of prime characteristic $p$ has a distinguished Scott module summand. In Green's theory of vertices and sources, this summand has vertex $Q$ where $Q \in \mathrm{Syl}_p(H)$ and trivial source. It is the unique summand of $M$ having the trivial module $F$ in its top and socle. In particular, $P_F$, the projective cover the trivial module, is the Scott module for trivial vertex.
Now let $\lambda$ be a partition of $n$ and let $M^\lambda$ be the Young permutation module for $FS_n$ induced from the Young subgroup $S_{\lambda_1} \times \cdots \times S_{\lambda_{\ell(\lambda)}}$. It is known that $M^\lambda$ has a distinguished summand $Y^\lambda$: this is the unique summand (in any given direct sum decomposition of $M^\lambda$) containing the Specht module $S^\lambda$. Moreover, $M^\lambda \cong Y^\lambda \oplus \bigoplus_{\mu > \lambda} c_\mu Y^\mu$ for some coefficients $c_\mu$.
Observe that $\lambda$ has a part of size $p$ or more if and only if $S_{\lambda_1} \times \cdots \times S_{\lambda_{\ell(\lambda)}}$ has a non-trivial Sylow $p$-subgroup and so if and only if the Scott module summand of $M^\lambda$ is non-projective. Therefore the lexicographically greatest partition $\mu$ such that $M^\mu$ has $P_F$ as a summand is the greatest partition with all parts $< p$, namely $((p-1)^s, t)$ where $0 \le t < p-1$ and $n = (p-1)s + t$. Moreover, since $P_F$ is not a summand of any $M^\lambda$ with $\lambda > \mu$ (they all have a Scott module summand, which cannot be project, so is not $P_F$), we have $Y^{((p-1)^s, t)} \cong P_F$.
This shows that the trivial module appears in the socle of the Specht module $S^{((p-1)^s,t)}$, and so using the basic result that $(S^\lambda)^\star \cong S^{\lambda'} \otimes \mathrm{sgn}$, the sign module appears at the top of $S^{((p-1)^s,t))'}$. Since $((p-1)^s,t)'$ is $p$-regular, this Specht module has a unique top composition factor and we get $D^{((p-1)^s, t)'} \cong \mathrm{sgn}$, as required. (And in agreement with the result from $p$-regularisation.)
