The problem is hard if $k$ is part of the input (not fixed):
Lemma 1. The problem is NP-hard, even for $n=2$.
The problem is also hard if $k$ (but not $n$) is fixed (with $k\ge 3$):
Lemma 2. The problem is strongly NP-hard, even for $k=3$.
The problem is in P if $k=1$ or $n=1$:
Lemma 3. The problem is in P for $k=1$ or $n=1$.
We leave open whether the problem is NP-hard for $k=2$.
Proof of Lemma 1. The proof is by reduction from the NP-hard Two-Way Balanced Partition problem.
Fix an instance $x \in \mathbb N^{2n'}$ of Two-Way Balanced Partition.
The goal is to determine whether there is a subset $S\subseteq [2n']$
of size $n'$
such that $\sum_{j\in S} x_j = \lambda$,
where $\lambda = \sum_{j=1}^{2n'} x_j / 2$.
Given $x$, the reduction outputs the instance $(n, k, w, \lambda)$ of the decision version of OP's problem
$n=2$, $k=n'$, and $w_{ij} = x_j$
for $(i, j) \in [2]\times [2n']$.
Next we verify that the reduction is correct.
Suppose that there is a feasible subset $S$ for $x$,
so $\sum_{j\in S} x_j = \lambda$.
Then the 1-to-$k$ matching $M$ defined by
$M(1) = S$ and $M(2) = [2n']\setminus S$
has cost $\lambda$.
Conversely, suppose that there is a 1-to-$k$ matching $M$
of cost at most $\lambda$.
Then taking $S=M(1)$ gives a feasible subset for $x$. $~~~\Box$
Proof of Lemma 2. The proof is by reduction from the strongly NP-hard 3-Partition problem.
Fix an instance $x \in \mathbb N^{3n}$ of 3-Partition.
The goal is to determine whether there is a partition $C$ of $[3n]$
into $n$ triples such that $\sum_{j\in C_i} x_j$ equals $\lambda$
for each triple $C_i$ in $C$,
where $\lambda = \sum_{j=1}^{3n} x_j/n$.
Given $x$, the reduction outputs the instance $(n, k, w, \lambda)$ of the decision version of OP's problem where
$k=3$, and $w_{ij} = x_j$
for $(i, j) \in [n]\times [3n]$.
Next we verify that the reduction is correct.
Suppose that there is a feasible partition $C$ for $x$,
so $\sum_{j\in C_i} x_j = \lambda$ for each triple $C_i$ in $C$.
Then the 1-to-3 matching $M$ defined by
$M(i) = C_i$ for each $i\in [n]$
has cost $\lambda$.
Conversely, suppose that there is a 1-to-3 matching $M$
of cost at most $\lambda$.
Then taking $C_i=M(i)$ for $i\in [n]$ gives a feasible partition $C$ for $x$. $~~~\Box$
Proof of Lemma 3. For any instance with $n=1$, there is only one possible solution $M(1) = [k]$, so the optimum can be found in linear time.
Given an instance $(k=1, n, w, \lambda)$ of the decision variant of OP's problem, the instance can be solved by reduction to the bipartite matching instance $G=(U, W, E)$ where $U = W = [n]$ and $E = \{(i, j) : w_{ij} \le \lambda\}$, which has a perfect matching iff OP's instance has a 1-to-1 matching of cost at most $\lambda$.
The optimization version of OP's problem can be solved using $O(\log n)$ iterations of binary search, over $\lambda$ in $\{w_{ij} : (i, j) \in [n] \times [n]\}$, to find the minimum $\lambda$ such that there is a solution of cost at most $\lambda$.
EDIT: The case $k=1$ was asked about here. (Answers pointed out that it is studied under the name Bottleneck Matching.)
$~~~~\Box$
