The lower bound can be substantially improved to $(1/\sqrt{2}-o(1))n$.

The proof goes in three parts: first a continuous analogue, then a LP analogue, and at last, the original problem.

The continuous analogue

The continuous analogue of the original problem can be stated as follows:

Let $U,D,R,L$ be Borel functions on $[0,1]\times [0,1] \rightarrow [0,1]$ satisfying $U+D+R+L=1$ and for every $(x,y) \in [0,1]\times [0,1]$, the value $\underset{0}{\overset{y}{\int}}U(x,t)dt+\underset{y}{\overset{1}{\int}}D(x,t)dt+\underset{0}{\overset{x}{\int}}R(t,y)dt+\underset{x}{\overset{1}{\int}}L(t,y)dt$ is at least $1/\sqrt2$.

Here $U,D,R,L$ are "densities of the up, down, right and left arrows", the first condition says that the "densities" at every point should add up to $1$, and the second condition says that every point should have at least a "mass" of $1/\sqrt{2}$ pointing to it.

The solution is as follows: Let $U$ be the function shown below.

The value $0$ is taken whenever $y \geq \frac12$.

When $y < \frac12 $, the value $\frac{2+\sqrt{2}}{4}$ is taken in the two side triangles, and the value $\frac{2-\sqrt{2}}{4}$ is taken in the lower triangle, and the value $\frac{1}{2}$ is taken otherwise (i.e. crossing dotted lines does not change the value of $U$). The top vertex of the lower triangle has coordinates $(\frac{1}{2},\frac{\sqrt{2}-1}{2})$, making the function $U$ left-right symmetric.

The functions $D, L, R$ are defined by their respective rotations of $U$, i.e. $D(x,y)=U(x,1-y)$, $R(x,y)=U(y,x)$ and $L(x,y)=R(1-x,y)$. It is easy to see that $U+D+L+R=1$.

Let $U^*(x,y)=\underset{0}{\overset{y}{\int}}U(x,t)dt$, i.e. the "mass" of up arrows pointing to $(x,y)$. Define the functions $D^*, L^*$ and $R^*$ in similar manners. Then the original problem reduces to studying the minimum of $U^*+D^*+R^*+L^*$.

The function $U^*$ is piecewise linear, and so is $U^*+D^*+R^*+L^*$.

The function $U^*+D^*+R^*+L^*$ is plotted below:

The numbers at the vertices represents the value of the function at that point. As the function is piecewise linear in every region, the function is determined by the number at the vertices. So it has minimum value $\frac{\sqrt{2}}{2}$. Thus the continuous problem is solved.

The LP analogue

Let $U',D',R',L'$ be $N\times N$ arrays with indices from $1$ to $N$ whose elements are in $[0,1]$. Now the respective value needed to be studied for the cell $(p,q)$ is $\underset{t=1}{\overset{q-1}{\sum}}U_{p,t}'+\underset{t=q+1}{\overset{N}{\sum}}D'_{p,t}+\underset{t=1}{\overset{p-1}{\sum}}R'_{t,q}+\underset{t=p+1}{\overset{N}{\sum}}L'_{t,q}$. This is the linear programming analogue of the original problem as noted by RobPratt.

The continuous problem provides a way to (approximately) solve the LP problem: Let $U_{p,q}'$ be the average of $U$ in $[\frac{p-1}{N},\frac{p}{N}] \times [\frac{q-1}{N},\frac{q}{N}]$ and define $D', R'$ and $L'$ similarly. Under such an assignment, the value for the cell $(p,q)$ is at least $\frac{\sqrt{2}}{2}N-C$ where $C$ is a constant independent with $N,p$ and $q$.

The arrays should be written in the order where $p$ goes from left to right, and $q$ goes from down to up. Only in such order can the meanings of $U', D', R'$ and $L'$ be retained.

The original problem

This is just a slight paraphrase of my comment in RobPratt's answer:

"Actually LP solutions serve as lower bounds. If we have a LP solution for $N \times N$ whose value is $K$, we can generate random arrow fillings on a $MN \times MN$ that matches the LP value arbitrarily close for sufficiently large $M$.

Dissect the $MN \times MN$ square into $N^2$ $M \times M$ sub-squares, corresponding to cells of the $N \times N$ LP solution. In a sub-square, let the directions of the cells be i.i.d. random variables with P(the arrow in the cell points at direction $x$) ($x\in\{\uparrow, \downarrow, \leftarrow, \rightarrow\}$) at least the value of the direction $x$ in the LP solution. Then as $M \rightarrow \infty$, it's asymptotically almost surely that every cell in the $MN \times MN$ square is pointed by at least $M(K-\epsilon)$ arrows, for any fixed $\epsilon>0$."

Taking into consideration that $K=\frac{\sqrt{2}}{2}n-O(1)$, we have proved that for every $\epsilon>0$ and sufficiently large $n$, there exists genuine solutions where every cell is pointed by at least $(\frac{\sqrt{2}}{2}-\epsilon)n$ arrows. (The divisibility issues are immaterial for any positive $\epsilon$.)

Hence $(1/\sqrt{2}-o(1))n$.

Discussion

By using dithering techniques, it may be possible to get a solution directly from the continuous analogue, skipping over the LP analogue, that has every cell pointed by at least $\frac{\sqrt{2}}{2}n-c$ arrows for a fixed constant $c>0$. But I have no references about the mathematical properties of dithering, so I do not know how to work out a proof.