Let $a$, $b$, $c$ and $d$ be integers with $0\leq a,b,c,d\leq 9$ and $a\neq0$, such that
$$(a+b+c+d)(a^2+b^2+c^2+d^2)^2=10^3a+10^2b+10c+d.$$
For brevity, I'll use your notation $N=SX^2$, where
\begin{eqnarray}
S&=&a+b+c+d,\\
X&=&a^2+b^2+c^2+d^2,\\
N&=&10^3a+10^2b+10c+d.
\end{eqnarray}
Observation 1: We have $a,b,c,d\leq5$ and $\min\{a,b,c,d\}\leq2$.
Proof. For $m:=\min\{a,b,c,d\}$ we have $S\geq4m$ and $X\geq4m^2$ and so
$$9999\geq N=SX^2\geq(4m)(4m^2)^2=64m^5,$$
from which it follows that $m\leq2$, so $N\leq9992$. Similarly, for $n:=\max\{a,b,c,d\}$ we have
$$9992\geq N=SX^2>(n)(n^2)^2=n^5,$$
from which it follows that $n\leq6$. But then $N\leq6662$ and so as before
$$6662\geq N=SX^2>(n)(n^2)^2=n^5,$$
from which it follows that $n\leq 5$. That is to say $0\leq a,b,c,d\leq5$ and $a\neq0$ and $\min\{a,b,c,d\}\leq2$. $\square$
Note that if $a\leq2$ then $N\leq2555$ and then the same argument shows that $n\leq4$.
Observation 2: We have $X\leq31$.
Proof. You have already found by repeated use of Cauchy-Schwarz that
$$S^2\leq 4X,\qquad N^2\leq 1010101X,\qquad X\leq S^2.$$
from which it follows that
$$X^5\leq S^2X^4=N^2\leq1010101X,$$
and hence that $X\leq31$. $\square$
This means that at most one of $a$, $b$, $c$ and $d$ is greater than $3$.
Observation 3: We have $a,b,c,d\leq4$.
Proof. If one of $a$, $b$, $c$ or $d$ equals $5$, then the others are no greater than $2$ because $X\leq31$. We already noted that if $a\leq2$ then $b,c,d\leq4$, a contradiction, so we must have $a=5$ and $b,c,d\leq2$. If one of them equals $2$ then
$$SX^2\geq(5+2+0+0)(5^2+2^2+0^2+0^2)^2=5887,$$
a contradiction, so $b,c,d\leq1$. Then $S=5+k$ and $X=25+k$ where $k=b+c+d$, and so
$$5000\leq(5+k)(25+k)^2\leq5111,$$
implies that $k=2$ yielding $N=SX=5103$, a contradiction. $\square$
Observation 4: We have $X\not\equiv0\pmod{3}$.
Proof. Suppose toward a contradiction that $X\equiv0\pmod{3}$. Then the identity
$$9(111a+11b+c)=N-S=SX^2-S=S(X+1)(X-1),$$
shows that $S\equiv0\pmod{9}$. By observation 3 we have $1\leq S\leq16$ and so $S=9$. Then up to permutation $(a,b,c,d)$ is one of the following tuples (keeping in mind that $X\equiv0\pmod{3}$):
$$(4,3,1,1),\qquad(3,3,3,0),\qquad(3,2,2,2).$$
But the corresponding values of $N$ are
\begin{eqnarray}
N&=&SX^2&=&(4+3+1+1)(4^2+3^2+1^2+1^2)^2&=&6561,\\
N&=&SX^2&=&(3+3+3+0)(3^2+3^2+3^2+0^2)^2&=&6561,\\
N&=&SX^2&=&(3+2+2+2)(3^2+2^2+2^2+2^2)^2&=&3969,\\
\end{eqnarray}
so these are not solutions. We conclude that $X\not\equiv0\pmod{3}$.$\square$
Observation 5: We have $a,b,c,d\leq3$ unless $N=2400$.
Proof. Order the digits from largest to smallest as $t\geq u\geq v\geq w$. Suppose that $t=4$. Then $X\leq31$ implies that $u\leq3$, and so $N\leq4300$. If $u=3$ or $u=v=2$ then
\begin{eqnarray}
N&=&SX^2\geq(4+3+0+0)(4^2+3^2+0^2+0^2)^2=4375,\\
N&=&SX^2\geq(4+2+2+0)(4^2+2^2+2^2+0^2)^2=4608,
\end{eqnarray}
both of which are impossible. Then $u\leq 2$ and $v,w\leq1$ and so
$$N=SX^2\leq(4+2+1+1)(4^2+2^2+1^2+1^2)=3872,$$
from which it follows that $N\leq2411$. On the other hand $u>0$ and so
$$N=SX^2\geq(4+1+0+0)(4^2+1^2+0^2+0^2)^2=1445,$$
from which it follows that $N\geq2004$, and so $u=2$. Keeping in mind that $X\not\equiv0\pmod{3}$ we find that either $v=w=0$ or $v=w=1$. A quick check shows that
\begin{eqnarray}
(2+4+1+1)(2^2+4^2+1^2+1^2)^2=3872,\\
(2+4+0+0)(2^2+4^2+0^2+0^2)^2=2400,
\end{eqnarray}
yielding the solution $N=2400$ as the only solution without $a,b,c,d\leq3$.$\square$
