First, we can evaluate the sum across the first row, which gives $(n+1)+1+(n-1)=2n+1$.  Evaluate the sum of the entries across the second row, $3+(2n-9)+n=3n-6$. Now, since we have a magic square, these two sums are equal.  So $2n+1=3n-6$. Isolating $n$, we obtain $n = \boxed{7}$.

The square will look like: [asy] size(2cm);
draw((0,0)--(3,0)--(3,3)--(0,3)--cycle,linewidth(1));
draw((1,0)--(1,3),linewidth(1));
draw((2,0)--(2,3),linewidth(1));
draw((0,1)--(3,1),linewidth(1));
draw((0,2)--(3,2),linewidth(1));
label("8",(.5,2.5));
label("1",(1.5,2.5));
label("6",(2.5,2.5));
label("3",(.5,1.5));
label("5",(1.5,1.5));
label("7",(2.5,1.5));
label("4",(.5,.5));
label("9",(1.5,.5));
label("2",(2.5,.5));
[/asy]
Question:
Below is a magic square, meaning that the sum of the numbers in each row, in each column, and in each of the $2$ main diagonals are equal. What is the value of $n$?

[asy]size(125);
for(int i = 0; i<4; ++i)
{

draw((0,i)--(3,i),linewidth(1));
}

for(int j = 0; j<4; ++j)
{

draw((j,0)--(j,3),linewidth(1));
}

label("$n-3$",(.5,.5));
label("3",(.5,1.5));
label("$n+1$",(.5,2.5));

label("$n+2$",(1.5,.5));
label("$2n-9$",(1.5,1.5));
label("$1$",(1.5,2.5));

label("$2$",(2.5,.5));
label("$n$",(2.5,1.5));
label("$n-1$",(2.5,2.5));
[/asy]
Answer:
Let's think step by step.

user
7
Below is a magic square, meaning that the sum of the numbers in each row, in each column, and in each of the $2$ main diagonals are equal. What is the value of $n$?

[asy]size(125);
for(int i = 0; i<4; ++i)
{

draw((0,i)--(3,i),linewidth(1));
}

for(int j = 0; j<4; ++j)
{

draw((j,0)--(j,3),linewidth(1));
}

label("$n-3$",(.5,.5));
label("3",(.5,1.5));
label("$n+1$",(.5,2.5));

label("$n+2$",(1.5,.5));
label("$2n-9$",(1.5,1.5));
label("$1$",(1.5,2.5));

label("$2$",(2.5,.5));
label("$n$",(2.5,1.5));
label("$n-1$",(2.5,2.5));
[/asy]
test