Is there a formula for the Frobenius-Schur indicator of a rep of a Lie group? Let $G$ be a simple algebraic group group over $\mathbb C$.
Let $V$ be a self-dual representation of $G$.
Let $\lambda$ be the highest weight of $V$.
Write $\lambda$ as a sum of fundamental weights: $\lambda=\sum \lambda_i\omega_i$ for $\lambda_i\in\mathbb N$.
Is there a simple formula for the Frobenius-Schur indicator of $V$ in terms of the numbers $\lambda_i$?

For the reader's convenience, I recall the definition of the Frobenius-Schur indicator. It is $1$ if the trivial rep occurs inside $Sym^2(V)$, and it is $-1$ if the trivial rep occurs inside $Alt^2(V)$.
 A: 
Proposition 7 in http://www.ams.org/journals/tran/1985-288-01/S0002-9947-1985-0773051-1/
contains a criterion (including a formula and proof; the proof is on page 131) to decide the type of the complex irreducible representation 
$\pi_\lambda$ of the compact connected semisimple Lie group $G$, where $\lambda$ is the highest weight, in terms of a maximal subset 
$\mathcal O=\{\beta_1,\ldots,\beta_\ell\}\subset\Delta^+$ of strongly orthogonal roots. Namely, $\pi_\lambda$ is unitary (not self-contragredient) if and only if 
$\lambda$ does not belong to the real span of $\mathcal O$. Otherwise
it is symplectic (resp. orthogonal), meaning that it leaves a quaternionic (resp. real) structure invariant, if and only if 
$$ k(\lambda) = \sum_{i=1}^\ell \frac{(\beta_i,\lambda)-(\beta_i,s_0\lambda)}{(\beta_i,\beta_i)} $$
is an odd (resp. even) integer. Here $(,)$ is the inner product defined from the Killing form, and 
$$s_0 := s_{\beta_1}\cdots s_{\beta_\ell}$$ 
is the Weyl group element that maps the Weyl chamber to its negative. 
The idea of the given proof is essentially to reduce the problem to a subgroup of $G$ isomorphic to the product of $\ell$ copies of $SU(2)$. 
The integer $k(\lambda)$ has an interesting meaning. Let $\mathcal U^k(\mathfrak g_\mathbb C)$ be the k-th level in the natural filtration of the universal
enveloping algebra of the complexified Lie algebra of $G$. 
Choose a highest weight vector $v_\lambda$ for $\pi_\lambda$. 
Then  $s_0v_\lambda \in\mathcal U^{k(\lambda)}(\mathfrak g_\mathbb C)v_\lambda$  but 
$s_0v_\lambda\not\in\mathcal U^{k'}(\mathfrak g_\mathbb C)\lambda$ for any $k'<k(\lambda)$. Here
$$ s_0v_\lambda = \pi(X_{-\beta_1})^{n_1}\cdots\pi(X_{-\beta_\ell})^{n_\ell}v_\lambda$$
where $$n_i=\frac{(\beta_i,\lambda)-(\beta_i,s_0\lambda)}{(\beta_i,\beta_i)} $$
and the $X_{-\beta_i}$ are root vectors. 
The table above lists the numbers $n_i$ for the fundamental representations of simple groups. 
A: As Jeff Adams indicates, there is a detailed proof in Bourbaki's Chapter 8 (originally published in French in 1975, but later translated into English).   See especially Table I at the end of Chapter 8 for a summary of all simple types in terms of their fundamental weights, referring back to $\S13$ for classical types and to $\S7$, no. 5, for exceptional types.  While the Onishchik-Vinberg book is quite useful as a learning tool, it is not a reasonable choice as a reference book.   Anyway, it's worthwhile to take a somewhat broader look at what is really going on here.
(1)  The term "Frobenius-Schur indicator" goes back to the joint 1906 paper by Frobenius and his student Schur in which they focused on a finite group $G$ and its complex irreducible characters.   Such a character $\chi$ arises from an irreducible matrix representation $\rho$ of $G$ over $\mathbb{C}$.    By the usual averaging argument, there is certainly a unitary representation equivalent to $\rho$ which affords $\chi$.   A natural  further question is whether there is a real representation of $G$ affording $\chi$.  Clearly the answer is no if $\chi$ takes nonreal values.  Then the original Frobenius-Schur indicator takes value 0, by definition. But even if $\chi$ is $\mathbb{R}$-valued, it may come from either a real orthogonal representation (then the indicator is written 1) or from a symplectic representation (then the indicator is $-1$).   A major step in the Frobenius-Schur work was a formula for the indicator computable from the character values: sum the values $\chi(g^2)$ over all $g \in G$, then average by dividing out $|G|$.
There are accounts of all this in many textbooks, such as those by Isaacs and by Serre (but don't start with their indexes; in Serre, try 13.2).     
(2) Though I haven't gone far into all the subsequent history, it's clear that the underlying question arises whenever you start with finite dimensional representations over $\mathbb{C}$ and ask what happens over $\mathbb{R}$.  In particular, the Frobenius-Schur question makes sense for complex (semi-)simple Lie groups and their finite dimensional representations parametrized by highest weights $\lambda$.   Here it is certainly possible to have indicator $=0$, when the underlying vector space $V$ admits no nondegenerate invariant bilinear form.   This case is avoided by assuming that the representation is self-dual: in case $V \cong V^*$ as a $G$-module, we get End$(V) \cong V^* \otimes V \cong V \otimes V$, so there is always a 1-dimensonal trivial $G$-submodule (corresponding to the scalar endomorphisms).   Since an irredudible $G$-module has at most one such nondegenerate form, it remains to decide whether it is orthogonal or symplectic.   
The combinatorics of roots and weights leads to the formal recipe indicated in a couple of the answers, which at first looks opaque.   But as Bourbaki points out, writing $m:=\sigma(\lambda)$ (in the notation used here), the half-integer $m/2$ has a natural interpretation as the sum of coefficients in the expression for $\lambda$ as a $\mathbb{Q}$-linear combination of simple roots.   Note in particular that when the root lattice equals the weight lattice (in types $G_2, F_4, E_8$), this half-integer is actually an integer and $m$ is even (so all indicators are 1).
(3) Generalizations in a number of directions have also been worked out, for example in the case of finite dimensional Hopf algebras (not just finite group algebras) or compact Lie groups or complex semisimple algebraic groups.   The formula discussed in the answers depends only on the root system, so it can be potentially applied over many kinds of fields including local fields and those of prime characteristic:   involutive maps are lurking everywhere. 
A: This is not a complete answer, but grew large for a comment.
Duality acts on the simple weights $\{\omega_i\}$, by $-w_0$ (where $w_0$ is the long element of the Weyl group), so we should really group them into the set of orbit sums
$$
D = \{\omega_i\,|\,\omega_i\,\, \text{is self-dual}\}\cup
\{\omega_j - w_0\cdot \omega_j\,|\,\omega_j\,\, \text{is not self-dual}\}.
$$
Then $\lambda = \sum_{d \in D} \lambda_d d$. Let $Q = \{d\in D\ :\ V_d$ is quaternionic$\}$. 
(Is it obvious that $Q$ doesn't contain any of the sums $\omega_j - w_0\cdot \omega_j$? I feel that should be true and obvious.)
It's easy to see that the indicator is multiplicative in $\lambda$, by tensoring the forms together, which leads to the formula $(-1)^{\sum_Q \lambda_q}$. (Of course Mikhail Borovoi's answer is of this form.)
A: Let us write the Frobenius-Schur indicator of the representation $V$ with highest weight $\lambda$
as $\mathrm{FS}(\lambda)$. 
Let $R$ denote the root system, and let $\Pi=\{\alpha_1,\dots,\alpha_l\}$ be a basis.
Let $\Pi^\vee=\{\alpha_1^\vee,\dots,\alpha_l^\vee\}$ denote the corresponding basis of the dual root system $R^\vee$.
Write $\lambda_i=\langle \lambda,\alpha_i^\vee\rangle$.
Exercise 4.3.12 on page 196 of the book by Onishchik and Vinberg "Lie Groups and Algebraic Groups", Springer-Verlag, 1990,  says that
$$\mathrm{FS}(\lambda)=(-1)^{\sigma(\lambda)},$$
where
$$\sigma(\lambda)=\sum_{i=1}^l r_i\lambda_i\tag{*} $$
and $r_i$ are the coordinates of of the sum of positive coroots $2\rho^\vee$ 
in the basis $\Pi^\vee$ of $R^\vee$.
No proof is given. 
The coordinates $r_i$ of $2\rho^\vee$ in the basis $\Pi^\vee$ 
can be found in the tables of Bourbaki "Groupes et  algèbres de Lie, Ch. 4,5,6",
formula (VII) in each table. Formulas for $r_i\ \mathrm{mod}\ 2$ 
are given in the last column of Table 3 on page 298 of the book by Onishchik and Vinberg and seem to agree with the formulas of Bourbaki.
For the reader's convenience I give below formulas for $\sigma(\lambda)$ following Table 3 of Onishchik and Vinberg, with the numbering of roots as in Bourbaki.
\begin{align*}
 A_l,\qquad &\sigma(\lambda)=\lambda_1+\lambda_3+\dots+\lambda_l\quad \text{for } l=2p+1,\text{ otherwise } 0;\\
 B_l,\qquad & \sigma(\lambda)=\lambda_l\quad\text{for } l=4q+1,\ l=4q+2,\text{ otherwise } 0;\\
 C_l\qquad &\sigma(\lambda)=\lambda_1+\lambda_3+\dots\quad\text{ for any } l;\\
 D_l\qquad &\sigma(\lambda)=\lambda_{l-1}+\lambda_l\quad\text{for }l=4q+2,\ l=4q+3,\text{ otherwise } 0;\\
 E_7\qquad &\sigma(\lambda)=\lambda_2+\lambda_5+\lambda_7\,.
\end{align*}
For all the other connected Dynkin diagrams, $\sigma(\lambda)=0$ for all $\lambda$ corresponding to self-dual representations.
Note that since we assume that $V$ is self-dual, the highest weight $\lambda$ is invariant under the involution $\nu$ of the Dynkin diagram that is the only nontrivial symmetry for $A_l$ ($l\ge 2$), $D_{2q+1}$ and $E_6$, and is trivial for all the other connected Dynkin diagrams. Taking this into account, we obtain that for $A_l$ the number $\sigma(\lambda)$ can be odd only for $l=4q+1$, and then we have
$\sigma(\lambda)\equiv\lambda_{(l+1)/2}\pmod{2}$, and that for $D_{4q+3}$ the number $\sigma(\lambda)$ is always even. Thus only $A_{4q+1}$, $B_{4q+1}$, $B_{4q+2}$, $C_l$, $D_{4q+2}$, and $E_7$ survive. This is what Exercise 4.3.13 in the book by Onishchik and Vinberg says. This agrees with the table in the answer of Claudio Gorodski.
A: The Frobenius-Schur indicator (of a self-dual finite dimensional representation) is 
$$
\chi_\pi(\exp(2\pi i\rho^\vee))
$$
where $\chi_\pi$ is the central character of $\pi$, $\rho^\vee$ is half the sum of the positive coroots, so $\exp(2\pi i\rho^\vee)$ is an element of order $2$ in the center of $G$. Equivalently if $\lambda$ is the highest weight then this equals
$$
e^{2\pi i\langle\lambda,\rho^\vee\rangle}
$$
This is, of course, equivalent to the other formulas cited, but is conceptually simpler. In particular: if $G$ is adjoint every (irreducible, finite dimensional, self-dual) representation is orthogonal.
See Bourbaki, Lie Groups and Lie Algebras, Chapters 7-9, Chapter IX, Section 7.2, Proposition 1. The proof is included, and is the one sketched by Borovoi. For a simpler proof based on the Tits group see this preprint.
