Closed form for the sequence defined by $a_0=1$ and $a_{n+1} = a_n + a_n^{-1}$

A closed form I doubt there is. But asymptotics are easy: $$ a_{n+1}^2=a_n^2+2+1/a_n^2, $$ hence, for every $n\ge1$, $$ a_n^2=2n+1+\sum_{k=0}^{n-1}\frac1{a_k^2}.\qquad\qquad\qquad\qquad (*) $$ This shows that $a_n^2\ge2n+2$ for every $n\ge1$, for example $a_{100}\ge\sqrt{202}>10\sqrt{2}>14$. In particular, $a_n\to+\infty$. Plugging this into $(*)$ yields $a_n^2=2n+1+o(n)$ hence $$ \sqrt{2n}\le a_n\le\sqrt{2n}+o(\sqrt{n}). $$ At this point, we know that $a_n^2\ge2n+2$ for every $n\ge1$. Using $(*)$ again, one sees that, for every $n\ge1$, $$ a_n^2\le2n+2+\sum_{k=1}^{n-1}\frac1{2+2k}\le2n+2+\frac12\log(n). $$ Which already shows that $$14.2<a_{100}<14.3$$ Plugging this upper bound of $a_n^2$ into $(*)$ would yield a refined lower bound of $a_n^2$. And one could then plug this refined lower bound into $(*)$ again to get a refined upper bound. And so on, back and forth between better and better upper bounds and better and better lower bounds. (No more asymptotics here.)

I agree, a closed form is very unlikely. As for more precise asymptotics, I think $a_n = \sqrt{2n} + 1/8\,{\frac {\sqrt {2}\ln \left( n \right) }{\sqrt {n}}}-{\frac {1}{ 128}}\,{\frac {\sqrt {2} \left( \ln \left( n \right) -2 \right) ^{2} + o(1)} {{n}^{3/2}}}$

To elaborate on how I got my answer: I started with @Did's $a(n) \approx \sqrt{2n}$ and looked for a next term. $a(n) = \sqrt{2n}$ would make $ a(n+1) - (a(n) + a(n)^{-1}) = \sqrt {2\,n+2}-\sqrt {2}\sqrt {n}-1/2\,{\frac {\sqrt {2}}{\sqrt {n}}} = - \frac{\sqrt{2}}{8} n^{-3/2} + O(n^{-5/2})$. With $a(n) = \sqrt{2n} + c n^{-1/2}$ I don't get a change in the $n^{-3/2}$ term, so I tried $a(n) = \sqrt{2n} + c \ln(n) n^{-1/2}$ and got $a(n+1) - (a(n) + a(n)^{-1}) = (-\frac{2}{\sqrt{8}} + c) n^{-3/2} + \ldots$. So to get rid of the $n^{-3/2}$ term I want $c = \frac{2}{\sqrt{8}}$. Then look at the leading term for $a(n) = \sqrt{2n} + \frac{2}{\sqrt{8}} \ln(n) n^{-1/2}$ and continue in that vein...