**Edit Jan 11, 2018:** One can make a little bit of progress by pushing my initial idea (pair partitions plus round robin) "to second order", i.e., make use of a little bit of cancellation. See further below.

**Edit Jan 10, 2018:**

*There is a flaw in the reasoning below as pointed out by Will.*

One needs more conditions on the combinatorial design like the only way to produce a pair partition of $[10]$ only using pairs in $\cup_i P_i$ is to take one the $P_i$'s. I don't know if this extra requirement can be satisfied.
Perhaps a round robin tournament scheduling algorithm might work.

**Initial version (faulty!):**

It follows from the existence of five set partitions $P_1,\ldots,P_5$ of the set $[10]:=\{1,2,\ldots,10\}$ into pairs such that these partitions are disjoint. Namely a pair $\{i,j\}\subset [10]$ with $i\neq j$ cannot be used by more than one partition.

Let us assume this existence. Then for each $P_i$ define its "standard permutation" as $\sigma_i$ given by listing the pairs by increasing order of its minimum and writing each pair in increasing order. Let $\epsilon_i$ be the sign of this standard permutation. We now define a tensor $T=(T_{i,j,k})_{(i,j,k)\in [10]\times[10]\times[5]}$ as follows.
If a pair $\{i,j\}\in P_k$ with $i<j$, set $T_{i,j,k}=\epsilon_k$ and $T_{j,i,k}=-\epsilon_k$. Otherwise make all the other tensor entries zero. Define the matrix of linear forms $M_{i,j}(x)=\sum_{k=1}^{5}T_{i,j,k} x_k$.
A moment of thought will show you that the Pfaffian of this matrix is a multiple of the Fermat quintic.

Fiddling around produces for instance this choice of partitions:

$$
P_1 =\{\ \{1,2\},\ \{3,4\},\ \{5,6\},\ \{7,8\},\ \{9,10\}\ \}
$$
$$
P_2 =\{\ \{1,10\},\ \{2,3\},\ \{4,5\},\ \{6,7\},\ \{8,9\}\ \}
$$
$$
P_3 =\{\ \{1,6\},\ \{2,7\},\ \{3,8\},\ \{4,9\},\ \{5,10\}\ \}
$$
$$
P_4 =\{\ \{1,3\},\ \{2,10\},\ \{4,6\},\ \{5,8\},\ \{7,9\}\ \}
$$
$$
P_5 =\{\ \{1,5\},\ \{2,6\},\ \{3,9\},\ \{4,7\},\ \{8,10\}\ \}
$$

**Edit Jan 11, 2018 (continued):**

*1) The Fermat quadratic* $F=x_1^2+x_2^2$.

My construction works as is and one can take for the partitions $P_1$ and $P_2$ any two of the three pair partitions of $[4]$.
To do this in a more principled way, let me modify the standard round robin scheduling algorithm in the even case.
Instead of fixing $1$ and turning everybody else, I will turn $1$ too. This gives the two arrays
$$
P_1=\left(
\begin{array}{cc}
1 & 2 \\
4 & 3
\end{array}
\right)\ \ ,\ \
P_2=\left(
\begin{array}{cc}
4 & 1 \\
3 & 2
\end{array}
\right)\ .
$$
The pairs correspond to the columns of the arrays.
The corresponding signs are $\epsilon_1=\epsilon_2=1$. This gives the matrix
$$
M=\left(
\begin{array}{cccc}
0 & x_2 & 0 & x_1 \\
-x_2 & 0 & x_1 & 0 \\
0 & -x_1 & 0 & x_2 \\
-x_1 & 0 & -x_2 & 0
\end{array}
\right)
$$
with ${\rm Pf}(M)=F$.

*2) The Fermat cubic* $F=x_1^3+x_2^3+x_3^3$.

The modified round robin rule gives the three pair partitions
$$
P_1=\left(
\begin{array}{ccc}
1 & 2 & 3 \\
6 & 5 & 4
\end{array}
\right)\ ,\
P_2=\left(
\begin{array}{ccc}
6 & 1 & 2 \\
5 & 4 & 3
\end{array}
\right)\ ,\
P_3=\left(
\begin{array}{ccc}
5 & 6 & 1 \\
4 & 3 & 2
\end{array}
\right)\ .
$$
One can also use an array of pairs to see what happens:
$$
\left(
\begin{array}{ccc}
(16) & (25) & (34) \\
(23) & (14) & (56) \\
(45) & (36) & (12)
\end{array}
\right)
$$
The partitions $P_1$, $P_2$, $P_3$ correspond to the rows of this array and their signs are $\epsilon_1=\epsilon_2=\epsilon_3=1$.

A rather tedious inspection shows that the only pair partitions one can form using any pairs in this $3\times 3$ array
are the previous ones and the three new ones corresponding to the columns of the array, i.e., $P_4=(16)(23)(45)$,
$P_5=(25)(14)(36)$ and $P_6=(34)(56)(12)$. Remark that this looks nicer if one draws chord diagrams on a circle with three colors for $P_1$, $P_2$, $P_3$. The corresponding signs are $\epsilon_4=1$, $\epsilon_5=-1$ and $\epsilon_6=1$.
Without any modification the Pfaffian should be
$$
x_1^3+x_2^3+x_3^3+x_1 x_2 x_3-x_1 x_2 x_3+x_1 x_2 x_3
$$
where I emphasized the individual contributions of $P_1,\ldots,P_6$ on purpose.
Now modify the weights of the pairs in the first row by factors $a,b,c$.
The Pfaffian now becomes
$$
abcx_1^3+x_2^3+x_3^3+ax_1 x_2 x_3-bx_1 x_2 x_3+cx_1 x_2 x_3
$$
so we get the Fermat cubic if we pick these weights so that
$$
abc=1\ \ ,\ \ a-b+c=0
$$
For example we can take $a=\theta$, $b=1$, $c=\theta^5$ where $\theta=e^{\frac{i\pi}{3}}$.

More explicitly, the Fermat cubic is $F={\rm Pf}(M)$
with
$$
M=\left(
\begin{array}{cccccc}
0 & x_3 & 0 & x_2 & 0 & \theta x_1 \\
-x_3 & 0 & x_2 & 0 & x_1 & 0 \\
0 & -x_2 & 0 & \theta^5 x_1 & 0 & x_3 \\
-x_2 & 0 & -\theta^5 x_1 & 0 & x_3 & 0 \\
0 & -x_1 & 0 & -x_3 & 0 & x_2 \\
-\theta x_1 & 0 & -x_3 & 0 & -x_2 & 0
\end{array}
\right)\ .
$$

There is an obvious pattern here but I don't know if it works for $n=4$ or the OP's question, that is, $n=5$. In fact, I just realized (via performing the same permutation on rows and columns) these are determinantal cases as in the comment by Will.